College Chemistry Quiz: Units Dimensional Analysis And Significant Figures
20 questions · exam conditions
0:00
Units Dimensional Analysis And Significant FiguresQuestion 1 of 20

Convert 2.50 × 10⁻³ kg to milligrams using dimensional analysis. What is the result?

2.50 × 10⁻⁶ mg
2.50 × 10⁻¹ mg
2.50 × 10³ mg
2.50 × 10⁶ mg
2.50 mg
← Back to quizzes

College Chemistry Quiz

College Chemistry Quiz: Units Dimensional Analysis And Significant Figures

Practice Units Dimensional Analysis And Significant Figures in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Units Dimensional Analysis And Significant Figures, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Convert 2.50 × 10⁻³ kg to milligrams using dimensional analysis. What is the result?

  1. 2.50 × 10⁻⁶ mg
  2. 2.50 × 10⁻¹ mg
  3. 2.50 × 10³ mg (correct answer)
  4. 2.50 × 10⁶ mg
  5. 2.50 mg
Explanation: Unit conversion problems test your ability to use dimensional analysis, a systematic method for converting between different units by multiplying by conversion factors that equal 1. To convert 2.50 × 10⁻³ kg to milligrams, you need to set up conversion factors that will cancel out kilograms and leave you with milligrams. Start with the given value and multiply by conversion factors: 2.50×10−3 kg×1000 g1 kg×1000 mg1 g2.50 \times 10^{-3} \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} \times \frac{1000 \text{ mg}}{1 \text{ g}} The kilograms cancel with the first fraction, and the grams cancel with the second fraction, leaving only milligrams: 2.50×10−3×1000×1000=2.50×10−3×106=2.50×103 mg2.50 \times 10^{-3} \times 1000 \times 1000 = 2.50 \times 10^{-3} \times 10^6 = 2.50 \times 10^3 \text{ mg} This confirms answer C is correct. Answer A (2.50 × 10⁻⁶ mg) results from incorrectly dividing by the conversion factors instead of multiplying, or confusing the direction of conversion. Answer B (2.50 × 10⁻¹ mg) comes from using only one conversion factor (kg to g) and forgetting the second step (g to mg). Answer D (2.50 × 10⁶ mg) happens when you multiply by an extra factor of 1000, possibly by converting kg directly to mg using 10⁶ instead of the step-by-step approach. Remember: always write out your conversion factors as fractions and verify that unwanted units cancel out completely. This visual check prevents most dimensional analysis errors.

Question 2

A student records a temperature measurement as 298.15 K. When converting this to Celsius using the relationship °C = K - 273.15, what is the result with the correct significant figures?

  1. 25 °C
  2. 25.0 °C
  3. 25.00 °C (correct answer)
  4. 25.000 °C
  5. 20 °C
Explanation: When you encounter temperature conversion problems, you're not just doing arithmetic—you're applying significant figures rules to maintain measurement precision throughout your calculation. To convert 298.15 K to Celsius, you subtract 273.15: 298.15−273.15=25.00298.15 - 273.15 = 25.00. The key insight is recognizing that both numbers have exactly the same number of decimal places (two), so your answer should maintain this precision. In significant figures, the rule for addition and subtraction is that your result should have the same number of decimal places as the measurement with the fewest decimal places. Since both 298.15 K and 273.15 have two decimal places, your answer must also show two decimal places, giving you 25.00 °C. Answer A (25 °C) incorrectly drops all decimal places, suggesting the temperature is only known to the nearest degree, which loses the precision of your original measurement. Answer B (25.0 °C) shows only one decimal place, which again understates the precision you actually have. Answer D (25.000 °C) incorrectly suggests you have three decimal places of precision, which would be claiming greater accuracy than your original data supports. Study tip: For temperature conversions, always check that your final answer maintains the same decimal precision as your starting measurement. The conversion factors (like 273.15) are exact relationships, so they don't limit your precision—your original measurement does.

Question 3

Convert 3.7 × 10⁻⁵ m³ to microliters (μL). Note that 1 m³ = 10⁶ L and 1 L = 10⁶ μL.

  1. 3.7 × 10⁻⁵ μL
  2. 3.7 × 10¹ μL
  3. 3.7 × 10⁴ μL
  4. 3.7 × 10⁷ μL (correct answer)
  5. 3.7 × 10¹⁰ μL
Explanation: Unit conversion problems in chemistry require systematic use of conversion factors to bridge between different scales of measurement. When you encounter multiple unit changes like this, the key is to set up a chain of conversions that cancels units properly. To convert 3.7×10−5 m33.7 \times 10^{-5} \text{ m}^3 to microliters, you need two conversion steps. First, convert cubic meters to liters using 1 m3=106 L1 \text{ m}^3 = 10^6 \text{ L}: 3.7×10−5 m3×106 L1 m3=3.7×10−5+6 L=3.7×101 L3.7 \times 10^{-5} \text{ m}^3 \times \frac{10^6 \text{ L}}{1 \text{ m}^3} = 3.7 \times 10^{-5+6} \text{ L} = 3.7 \times 10^1 \text{ L} Next, convert liters to microliters using 1 L=106 μL1 \text{ L} = 10^6 \text{ μL}: 3.7×101 L×106 μL1 L=3.7×101+6 μL=3.7×107 μL3.7 \times 10^1 \text{ L} \times \frac{10^6 \text{ μL}}{1 \text{ L}} = 3.7 \times 10^{1+6} \text{ μL} = 3.7 \times 10^7 \text{ μL} This confirms answer choice D is correct. Answer choice A (3.7×10−53.7 \times 10^{-5} μL) represents no conversion at all—just changing the units without applying conversion factors. Answer choice B (3.7×1013.7 \times 10^1 μL) shows only the first conversion step completed, stopping at liters instead of continuing to microliters. Answer choice C (3.7×1043.7 \times 10^4 μL) likely results from incorrectly using 10310^3 instead of 10610^6 for one of the conversion factors. Study tip: Always write out your conversion factors as fractions and verify that units cancel completely. When working with scientific notation, add exponents when multiplying—this makes tracking the math much easier and prevents calculation errors.

Question 4

Express the result of 12.345 + 1.2 + 0.06 with the correct number of significant figures according to addition rules.

  1. 13.605
  2. 13.60
  3. 13.6 (correct answer)
  4. 13
  5. 14
Explanation: When working with significant figures in addition and subtraction, you need to focus on decimal places, not the total number of significant figures. The rule is simple: your answer can only be as precise as the least precise measurement in terms of decimal places. Let's examine each number in the calculation 12.345+1.2+0.0612.345 + 1.2 + 0.06:
  • 12.34512.345 has 3 decimal places
  • 1.21.2 has 1 decimal place
  • 0.060.06 has 2 decimal places
The limiting factor is 1.21.2 with only 1 decimal place. This means your final answer must be rounded to 1 decimal place. Performing the calculation: 12.345+1.2+0.06=13.60512.345 + 1.2 + 0.06 = 13.605 Now rounding to 1 decimal place: 13.613.6 Answer A (13.605) shows the raw calculation without applying significant figure rules—this ignores the precision limitations of your measurements. Answer B (13.60) incorrectly keeps 2 decimal places, likely confused by the 0.060.06 term. Answer D (13) rounds to the nearest whole number, which would only be correct if one of your measurements was given as a whole number with no decimal point. Study tip: In addition/subtraction problems, always identify the measurement with the fewest decimal places first—that determines your final answer's precision. Don't get distracted by counting total significant figures, which only matters for multiplication and division.

Question 5

A student needs to prepare 250.0 mL of 0.100 M NaCl solution from solid NaCl (molar mass = 58.44 g/mol). What mass of NaCl is required, expressed with appropriate significant figures?

  1. 1.46 g (correct answer)
  2. 1.461 g
  3. 1.4610 g
  4. 1.46100 g
  5. 1.5 g
Explanation: When you encounter solution preparation problems, you're working with molarity calculations that require careful attention to significant figures. The key relationship is: Molarity=moles of soluteliters of solution\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} To find the required mass of NaCl, first calculate the moles needed. With 0.250 L of 0.100 M solution: moles NaCl=0.100 M×0.250 L=0.0250 mol\text{moles NaCl} = 0.100 \text{ M} \times 0.250 \text{ L} = 0.0250 \text{ mol} Next, convert moles to grams using the molar mass: mass=0.0250 mol×58.44 g/mol=1.461 g\text{mass} = 0.0250 \text{ mol} \times 58.44 \text{ g/mol} = 1.461 \text{ g} Now apply significant figure rules. Your given data includes 250.0 mL (4 sig figs), 0.100 M (3 sig figs), and 58.44 g/mol (4 sig figs). The limiting factor is 0.100 M with 3 significant figures, so your final answer must have 3 significant figures. Choice A (1.46 g) correctly rounds 1.461 g to 3 significant figures. Choice B (1.461 g) shows 4 significant figures, which exceeds what your data justifies. Choice C (1.4610 g) has 5 significant figures and inappropriately adds a trailing zero. Choice D (1.46100 g) extends to 6 significant figures with multiple unjustified trailing zeros. Remember: your final answer can never be more precise than your least precise measurement. In molarity calculations, always identify which piece of given data has the fewest significant figures—that determines your final answer's precision, regardless of how many decimal places your calculator displays.

Question 6

Convert 75.0 mmHg to pascals using the conversion factor 1 atm = 760 mmHg and 1 atm = 101,325 Pa. Express the answer with correct significant figures.

  1. 9.99×1039.99 \times 10^3 Pa
  2. 9.993×1039.993 \times 10^3 Pa
  3. 1.00×1041.00 \times 10^4 Pa (correct answer)
  4. 9.99342×1039.99342 \times 10^3 Pa
  5. 1.0×1041.0 \times 10^4 Pa
Explanation: Unit conversion problems like this test your ability to use dimensional analysis and apply significant figures rules correctly. When converting between pressure units, you need to set up conversion factors that cancel unwanted units and leave you with the desired units. To convert 75.0 mmHg to pascals, you can use the given conversion factors in sequence: 75.0 mmHg×1 atm760 mmHg×101,325 Pa1 atm=9,993.4... Pa75.0 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} \times \frac{101,325 \text{ Pa}}{1 \text{ atm}} = 9,993.4... \text{ Pa} The key insight is applying significant figures correctly. Your initial measurement (75.0 mmHg) has three significant figures, so your final answer must also have three significant figures. The raw calculation gives 9,993.4 Pa, which rounds to 1.00×1041.00 \times 10^4 Pa when expressed with three significant figures. Looking at the wrong answers: Choice A (9.99×1039.99 \times 10^3 Pa) incorrectly rounds the raw result but maintains the wrong order of magnitude. Choice B (9.993×1039.993 \times 10^3 Pa) shows four significant figures, violating the three-significant-figure rule from the original measurement. Choice D (9.99342×1039.99342 \times 10^3 Pa) displays six significant figures, completely ignoring significant figures rules and showing false precision. Remember that in unit conversions, your final answer's precision is limited by the least precise measurement in the problem. Always count significant figures in your starting value and round your final answer accordingly, even if it means changing the order of magnitude in scientific notation.

Question 7

A student measures the following masses during an experiment: 5.43 g, 12.1 g, and 0.078 g. What is the sum expressed with the correct number of decimal places?

  1. 17.608 g
  2. 17.61 g
  3. 17.6 g (correct answer)
  4. 18 g
  5. 17.608000 g
Explanation: When you're adding measurements in chemistry, significant figures rules require you to consider the precision of each measurement, particularly the number of decimal places. For addition and subtraction, your final answer should have the same number of decimal places as the measurement with the fewest decimal places. Let's examine each measurement: 5.43 g has 2 decimal places, 12.1 g has 1 decimal place, and 0.078 g has 3 decimal places. The measurement with the fewest decimal places is 12.1 g with just 1 decimal place, so your final answer must be rounded to 1 decimal place. Adding the values: 5.43+12.1+0.078=17.6085.43 + 12.1 + 0.078 = 17.608 g Now round to 1 decimal place: 17.6 g, which is answer C. Answer A (17.608 g) shows the raw calculation without proper rounding—this ignores significant figures rules entirely. Answer B (17.61 g) rounds to 2 decimal places, which would be correct if you mistakenly used 5.43 g (2 decimal places) as your limiting measurement instead of 12.1 g. Answer D (18 g) rounds to the nearest whole number, which is too extreme—this would only be appropriate if one of your measurements had been given as a whole number. Remember this key distinction: for multiplication and division, count total significant figures, but for addition and subtraction, count decimal places. Always let the least precise measurement (fewest decimal places) determine your final answer's precision.

Question 8

Convert 4.5×10−84.5 \times 10^{-8} cm to nanometers (nm). Use the relationship 1 cm = 10−210^{-2} m and 1 nm = 10−910^{-9} m.

  1. 4.5×10−64.5 \times 10^{-6} nm
  2. 4.5×10−104.5 \times 10^{-10} nm
  3. 4.5×10−14.5 \times 10^{-1} nm (correct answer)
  4. 4.5×1014.5 \times 10^{1} nm
  5. 4.5×10−84.5 \times 10^{-8} nm
Explanation: Unit conversion problems in chemistry require systematic application of conversion factors to bridge between different measurement scales. When converting between units with scientific notation, you'll multiply by conversion factors and adjust the powers of 10. Starting with 4.5×10−84.5 \times 10^{-8} cm, you need to convert to nanometers using the given relationships. First, convert centimeters to meters: 4.5×10−8 cm×10−2 m1 cm=4.5×10−10 m4.5 \times 10^{-8} \text{ cm} \times \frac{10^{-2} \text{ m}}{1 \text{ cm}} = 4.5 \times 10^{-10} \text{ m} Next, convert meters to nanometers: 4.5×10−10 m×1 nm10−9 m=4.5×10−10×109 nm=4.5×10−1 nm4.5 \times 10^{-10} \text{ m} \times \frac{1 \text{ nm}}{10^{-9} \text{ m}} = 4.5 \times 10^{-10} \times 10^{9} \text{ nm} = 4.5 \times 10^{-1} \text{ nm} This confirms answer C is correct. Looking at the wrong answers: Answer A (4.5×10−64.5 \times 10^{-6} nm) results from incorrectly multiplying the exponents instead of adding them during the meter-to-nanometer conversion. Answer B (4.5×10−104.5 \times 10^{-10} nm) stops at the intermediate step—this is the value in meters, not nanometers. Answer D (4.5×1014.5 \times 10^{1} nm) comes from using incorrect conversion relationships or flipping the direction of the conversions. Study tip: Always set up conversion factors so unwanted units cancel out, and remember that dividing by 10−910^{-9} is the same as multiplying by 10910^{9}. Work systematically through each conversion step rather than trying to do everything at once.

Question 9

A measurement of 0.02030 g contains how many significant figures?

  1. 2
  2. 3
  3. 4 (correct answer)
  4. 5
  5. 6
Explanation: Significant figures questions test your understanding of measurement precision and which digits in a number are meaningful based on how the measurement was recorded. To count significant figures in 0.02030 g, you need to apply the key rules systematically. Leading zeros (the zeros before the first non-zero digit) are never significant—they're just placeholders that indicate the decimal point's position. So the "0.0" at the beginning doesn't count. The digits 2, 0, 3, and the final 0 are all significant. The 2 and 3 are obviously significant as non-zero digits. The middle 0 is significant because it falls between two non-zero digits. Most importantly, the trailing zero after the 3 is significant because it comes after the decimal point and after other significant digits, indicating the measurement was precise to that decimal place. This gives us 4 significant figures total. Choice (A) of 2 figures would ignore the zeros entirely, which misapplies the leading zero rule to all zeros. Choice (B) of 3 figures likely counts 2, 0, and 3 but incorrectly treats the final zero as insignificant. Choice (D) of 5 figures incorrectly counts one of the leading zeros as significant. Remember this pattern: in decimal numbers, zeros are significant when they're between non-zero digits or when they trail after both a decimal point and other significant digits. Leading zeros before the first non-zero digit never count. The correct answer is (C) 4 significant figures.

Question 10

A student is asked to determine the concentration of an unknown glucose solution using a calibration curve. The student prepares standard solutions and measures their absorbances. The data collected shows that absorbance = 0.0234 × concentration (g/L) + 0.003. For the unknown solution, the measured absorbance is 0.147.

Calculate the concentration of the unknown glucose solution in g/L with appropriate significant figures.

  1. 6.2 g/L (correct answer)
  2. 6.15 g/L
  3. 6.154 g/L
  4. 6.1538 g/L
  5. 6 g/L
Explanation: When you encounter calibration curve problems, you're applying Beer's Law concepts where absorbance relates linearly to concentration. The key is using the given equation correctly and applying significant figure rules. To find the unknown concentration, rearrange the calibration equation: absorbance = 0.0234 × concentration + 0.003. Solving for concentration: concentration=absorbance−0.0030.0234\text{concentration} = \frac{\text{absorbance} - 0.003}{0.0234} Substituting the measured absorbance of 0.147: concentration=0.147−0.0030.0234=0.1440.0234=6.1538... g/L\text{concentration} = \frac{0.147 - 0.003}{0.0234} = \frac{0.144}{0.0234} = 6.1538... \text{ g/L} Now comes the crucial step: significant figures. Your measured absorbance (0.147) has three significant figures, which limits your final answer. The calibration equation coefficients appear to have three significant figures as well. Therefore, your answer should be rounded to two decimal places, giving 6.2 g/L. Choice A (6.2 g/L) correctly applies significant figure rules. Choice B (6.15 g/L) shows an intermediate rounding step but still has too many significant figures for the given data precision. Choice C (6.154 g/L) and Choice D (6.1538 g/L) both report far more precision than the experimental data supports—a common mistake when students report all calculator digits without considering measurement uncertainty. Remember: in analytical chemistry problems, your final answer can never be more precise than your least precise measurement. Always check significant figures as your final step, especially when the answer choices differ only in decimal places.

Question 11

Calculate the result of (8.314 J mol−1K−1)×(298.15 K)×(0.50 mol)(8.314 \text{ J mol}^{-1} \text{K}^{-1}) \times (298.15 \text{ K}) \times (0.50 \text{ mol}) with appropriate significant figures.

  1. 1.2×1031.2 \times 10^3 J (correct answer)
  2. 1.24×1031.24 \times 10^3 J
  3. 1.239×1031.239 \times 10^3 J
  4. 1.2392×1031.2392 \times 10^3 J
  5. 1.23923×1031.23923 \times 10^3 J
Explanation: This problem tests your understanding of significant figures in calculations—a fundamental skill in chemistry where precision matters for experimental data and thermodynamic calculations. To solve this, you multiply the three values: (8.314)×(298.15)×(0.50)=1239.2271 J(8.314) \times (298.15) \times (0.50) = 1239.2271 \text{ J} The key is determining how many significant figures to report. When multiplying or dividing, your answer should have the same number of significant figures as the measurement with the fewest significant figures. Let's analyze each value:
  • 8.314 J mol⁻¹K⁻¹ has 4 significant figures
  • 298.15 K has 5 significant figures
  • 0.50 mol has 2 significant figures (the trailing zero after the decimal is significant)
Since 0.50 mol has only 2 significant figures, your final answer must be rounded to 2 significant figures: 1200 J, or 1.2×1031.2 \times 10^3 J. Answer A is correct with proper significant figures. Answer B (1.24×1031.24 \times 10^3) shows 3 significant figures, which is incorrect. Answer C (1.239×1031.239 \times 10^3) shows 4 significant figures, ignoring the limiting factor entirely. Answer D (1.2392×1031.2392 \times 10^3) shows 5 significant figures, which greatly exceeds what the data supports. Remember: in multiplication and division, you're only as precise as your least precise measurement. Always identify the limiting factor first, then perform the calculation and round accordingly. This prevents false precision in your chemical calculations.

Question 12

When measuring 25.0 mL of solution using a graduated cylinder, a student records the volume as 25.2 mL. Calculate the percent error in this measurement.

  1. 0.8% (correct answer)
  2. 0.79%
  3. 1.2%
  4. 8.0%
  5. -0.8%
Explanation: Percent error questions test your understanding of measurement accuracy and how to quantify the difference between measured and true values. When you encounter these problems, you need to identify the true value, the measured value, and apply the percent error formula. The percent error formula is: Percent Error=∣measured value−true value∣∣true value∣×100%\text{Percent Error} = \frac{|\text{measured value} - \text{true value}|}{|\text{true value}|} \times 100\% Here, the true value is 25.0 mL (what should have been measured) and the measured value is 25.2 mL (what was actually recorded). Substituting into the formula: Percent Error=∣25.2−25.0∣∣25.0∣×100%=0.225.0×100%=0.008×100%=0.8%\text{Percent Error} = \frac{|25.2 - 25.0|}{|25.0|} \times 100\% = \frac{0.2}{25.0} \times 100\% = 0.008 \times 100\% = 0.8\% This confirms answer A is correct. Looking at the wrong answers: B (0.79%) likely results from rounding errors or calculator mistakes during the division step. C (1.2%) suggests someone may have incorrectly used the measured value (25.2) as the denominator instead of the true value (25.0). D (8.0%) represents a decimal place error—moving the decimal one place too far right, turning 0.8% into 8.0%. Remember that percent error always uses the true (accepted) value in the denominator, not the measured value. Also, pay careful attention to significant figures and decimal placement in your final calculation. These problems frequently test mathematical precision alongside chemical measurement concepts.

Question 13

A reaction requires 0.0125 mol of reactant. If the molar mass of the reactant is 158.03 g/mol, what mass should be weighed out, expressed to the appropriate number of significant figures?

  1. 1.98 g (correct answer)
  2. 1.975 g
  3. 1.9754 g
  4. 1.98375 g
  5. 2.0 g
Explanation: When you encounter stoichiometry problems involving mass calculations, you're working with the fundamental relationship between moles and mass through molar mass. The key is not just getting the right numerical answer, but expressing it with the correct number of significant figures based on your given data. To find the mass needed, you multiply moles by molar mass: mass=moles×molar mass=0.0125 mol×158.03 g/mol=1.97538 g\text{mass} = \text{moles} \times \text{molar mass} = 0.0125 \text{ mol} \times 158.03 \text{ g/mol} = 1.97538 \text{ g} Now comes the critical part: significant figures. Your given values are 0.0125 mol (3 significant figures) and 158.03 g/mol (5 significant figures). The answer must be limited by the least precise measurement, which is 3 significant figures. Rounding 1.97538 g to 3 significant figures gives 1.98 g, making A correct. Option B (1.975 g) shows 4 significant figures, which exceeds what your data supports. Option C (1.9754 g) displays 5 significant figures—far too many given your starting precision. Option D (1.98375 g) represents 6 significant figures and appears to be the unrounded calculator result, ignoring significant figure rules entirely. Study tip: In stoichiometry calculations, always identify the number of significant figures in each given value before you start calculating. Your final answer can never be more precise than your least precise starting measurement. This principle applies across all quantitative chemistry problems and is frequently tested.

Question 14

A student measures the volume of a gas as 2.45 L at 22.0°C and 1.05 atm. Using the ideal gas law PV=nRTPV = nRT with R=0.08206R = 0.08206 L·atm·mol⁻¹·K⁻¹, calculate the number of moles present with appropriate significant figures.

  1. 0.106 mol (correct answer)
  2. 0.1064 mol
  3. 0.10644 mol
  4. 0.106439 mol
  5. 0.11 mol
Explanation: When you encounter ideal gas law problems, remember that this equation connects all the key gas properties, but you must always convert temperature to Kelvin and pay careful attention to significant figures in your final answer. To find the number of moles, rearrange the ideal gas law to solve for n: n=PVRTn = \frac{PV}{RT}. First, convert the temperature from Celsius to Kelvin: T=22.0°C+273.15=295.2KT = 22.0°C + 273.15 = 295.2 K. Now substitute the values: n=(1.05 atm)(2.45 L)(0.08206 L\cdotpatm\cdotpmol−1\cdotpK−1)(295.2 K)=2.572524.229=0.106439 moln = \frac{(1.05 \text{ atm})(2.45 \text{ L})}{(0.08206 \text{ L·atm·mol}^{-1}\text{·K}^{-1})(295.2 \text{ K})} = \frac{2.5725}{24.229} = 0.106439 \text{ mol} However, your final answer must reflect the appropriate number of significant figures. Looking at the given data: 2.45 L has 3 sig figs, 22.0°C has 3 sig figs, and 1.05 atm has 3 sig figs. Therefore, your answer should have 3 significant figures. Choice A (0.106 mol) correctly rounds to 3 significant figures. Choice B (0.1064 mol) shows 4 significant figures, which exceeds what the data supports. Choice C (0.10644 mol) shows 5 significant figures, and Choice D (0.106439 mol) shows 6 significant figures - both are inappropriately precise given the input data. The key strategy here is to always determine significant figures from your given data before performing calculations, then round your final numerical result accordingly. Many students get the math right but lose points for incorrect significant figures.

Question 15

Calculate (6.022×1023)×(1.66×10−24)(6.022 \times 10^{23}) \times (1.66 \times 10^{-24}) and express the result with appropriate significant figures.

  1. 1.0
  2. 1.00 (correct answer)
  3. 0.999652
  4. 1.000
  5. 9.9965×10−19.9965 \times 10^{-1}
Explanation: When you encounter calculations involving scientific notation and significant figures, you need to both perform the arithmetic correctly and apply sig fig rules properly. Let's calculate (6.022×1023)×(1.66×10−24)(6.022 \times 10^{23}) \times (1.66 \times 10^{-24}). First, multiply the coefficients: 6.022×1.66=9.996526.022 \times 1.66 = 9.99652. Then handle the powers of 10: 1023×10−24=10−1=0.110^{23} \times 10^{-24} = 10^{-1} = 0.1. So we get 9.99652×0.1=0.9996529.99652 \times 0.1 = 0.999652. Now comes the crucial part: significant figures. The first number (6.022) has 4 significant figures, while the second number (1.66) has only 3 significant figures. When multiplying, your answer can have no more significant figures than the measurement with the fewest sig figs. Therefore, you must round 0.999652 to 3 significant figures, giving you 1.00. Looking at the wrong answers: Choice A (1.0) shows only 2 significant figures, which undershoots the precision you should maintain. Choice C (0.999652) represents the raw calculation before applying sig fig rules—a common mistake where students forget to round appropriately. Choice D (1.000) has 4 significant figures, which exceeds what the least precise measurement (1.66) allows. The correct answer is B (1.00). Remember this key strategy: always identify the number of significant figures in each given value before calculating, then round your final answer to match the least precise measurement. Don't let the raw calculator result fool you into thinking more precision is justified than your data actually supports.

Question 16

A student performs a titration experiment to determine the concentration of an unknown HCl solution. The student uses 0.1050 M NaOH as the titrant and records the following data: Initial buret reading: 1.25 mL, Final buret reading: 26.78 mL, Volume of HCl solution: 25.00 mL.

Calculate the molarity of the HCl solution with the appropriate number of significant figures.

  1. 0.107 M (correct answer)
  2. 0.1070 M
  3. 0.10704 M
  4. 0.107044 M
  5. 0.11 M
Explanation: When you encounter acid-base titration problems, you're applying stoichiometry to determine unknown concentrations. The key is systematically working through the volume and molarity relationships. First, calculate the volume of NaOH used: 26.78 mL−1.25 mL=25.53 mL26.78 \text{ mL} - 1.25 \text{ mL} = 25.53 \text{ mL} Next, find moles of NaOH: 0.1050 M×0.02553 L=0.002681 mol NaOH0.1050 \text{ M} \times 0.02553 \text{ L} = 0.002681 \text{ mol NaOH} Since HCl and NaOH react in a 1:1 ratio (HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}), moles of HCl equals moles of NaOH. Finally, calculate HCl molarity: 0.002681 mol0.02500 L=0.107044 M\frac{0.002681 \text{ mol}}{0.02500 \text{ L}} = 0.107044 \text{ M} Now for significant figures: your measurements have 4, 4, 3, and 4 significant figures respectively. The limiting factor is the 25.0 mL volume with 3 significant figures, so your answer must have 3 significant figures. Choice A (0.107 M) correctly rounds to 3 significant figures. Choice B (0.1070 M) shows 4 significant figures, which exceeds what your data supports. Choice C (0.10704 M) shows 5 significant figures, and Choice D (0.107044 M) shows 6 significant figures—both are inappropriately precise given your measurements. Study tip: In titration calculations, always identify the measurement with the fewest significant figures first—this determines your final answer's precision. Many students calculate correctly but lose points by reporting too many or too few significant figures.

Question 17

Express 0.000456 in scientific notation with three significant figures.

  1. 4.56 × 10⁻⁴ (correct answer)
  2. 4.6 × 10⁻⁴
  3. 0.456 × 10⁻³
  4. 45.6 × 10⁻⁵
  5. 4.560 × 10⁻⁴
Explanation: Scientific notation expresses numbers as a product of a decimal between 1 and 10 multiplied by a power of 10. This format is essential in chemistry for handling very large or very small measurements while maintaining proper significant figures. To convert 0.000456 to scientific notation, you need to move the decimal point to create a number between 1 and 10. Starting with 0.000456, move the decimal point 4 places to the right to get 4.56. Since you moved the decimal point 4 places to the right, the exponent will be -4, giving you 4.56×10−44.56 \times 10^{-4}. The number 4.56 already has three significant figures (4, 5, and 6), so no rounding is needed. Looking at the incorrect options: Option B (4.6×10−44.6 \times 10^{-4}) incorrectly rounds to only two significant figures instead of the required three. Option C (0.456×10−30.456 \times 10^{-3}) has the wrong format because 0.456 is not between 1 and 10 – proper scientific notation requires the coefficient to be at least 1. Option D (45.6×10−545.6 \times 10^{-5}) also uses incorrect formatting since 45.6 is greater than 10, violating the fundamental rule of scientific notation. The correct answer is A: 4.56×10−44.56 \times 10^{-4}. Study tip: Remember that proper scientific notation always has exactly one non-zero digit before the decimal point. Count decimal places carefully when determining the exponent, and always check that your final answer maintains the correct number of significant figures.

Question 18

Round 0.008247 to three significant figures and express the result in proper scientific notation.

  1. 8.25×10−38.25 \times 10^{-3} (correct answer)
  2. 8.24×10−38.24 \times 10^{-3}
  3. 0.00825
  4. 8.2×10−38.2 \times 10^{-3}
  5. 0.824×10−20.824 \times 10^{-2}
Explanation: Significant figures and scientific notation are fundamental skills in chemistry because they communicate the precision of your measurements and calculations. When you encounter a problem asking you to round to a specific number of significant figures, you need to identify which digits count, apply rounding rules correctly, and express the result in proper scientific notation. Starting with 0.008247, the leading zeros don't count as significant figures—they're just placeholders. The significant digits are 8, 2, 4, and 7. To round to three significant figures, you keep the first three (8, 2, 4) and look at the fourth digit (7) to determine whether to round up or down. Since 7 ≥ 5, you round up, changing the 4 to 5. This gives you 8.25. In proper scientific notation, you express this as 8.25×10−38.25 \times 10^{-3}, where the decimal point is moved three places to the right from its original position. Choice A (8.25×10−38.25 \times 10^{-3}) correctly applies the rounding rule and scientific notation. Choice B (8.24×10−38.24 \times 10^{-3}) fails to round up when the fourth digit is 7. Choice C (0.00825) has the right digits but isn't in scientific notation as requested. Choice D (8.2×10−38.2 \times 10^{-3}) only shows two significant figures instead of the required three. Remember: when rounding, always look at the digit immediately after your target precision. If it's 5 or greater, round up; if it's less than 5, round down. Scientific notation should have exactly one non-zero digit before the decimal point.

Question 19

Calculate the result of (4.56 × 10²) × (2.1 × 10⁻⁴) ÷ (3.789 × 10⁻¹) with the appropriate significant figures.

  1. 2.5 × 10⁻¹ (correct answer)
  2. 2.53 × 10⁻¹
  3. 2.534 × 10⁻¹
  4. 0.25 × 10⁻¹
  5. 2.5 × 10⁻²
Explanation: When you encounter scientific notation problems involving multiplication and division, you need to handle both the coefficients and the exponents systematically while applying significant figures rules. First, let's work through the calculation step by step: (4.56×102)×(2.1×10−4)(3.789×10−1)\frac{(4.56 \times 10²) \times (2.1 \times 10⁻⁴)}{(3.789 \times 10⁻¹)} Multiply the coefficients: 4.56×2.1=9.5764.56 \times 2.1 = 9.576 Add exponents for multiplication: 102+(−4)=10−210^{2 + (-4)} = 10^{-2} So the numerator becomes: 9.576×10−29.576 \times 10^{-2} Now divide by the denominator: Coefficients: 9.5763.789=2.527...\frac{9.576}{3.789} = 2.527... Subtract exponents: 10−2−(−1)=10−110^{-2 - (-1)} = 10^{-1} This gives us 2.527×10−12.527 \times 10^{-1} Now apply significant figures rules. In multiplication and division, your answer should have the same number of significant figures as the measurement with the fewest significant figures. Here, 2.1 has only 2 significant figures, so round to 2 significant figures: 2.5×10−12.5 \times 10^{-1} Answer A (2.5×10−12.5 \times 10^{-1}) is correct. Answer B (2.53×10−12.53 \times 10^{-1}) shows 3 significant figures, which violates the rule. Answer C (2.534×10−12.534 \times 10^{-1}) shows 4 significant figures, also incorrect. Answer D (0.25×10−10.25 \times 10^{-1}) has the wrong coefficient and isn't in proper scientific notation form. Remember: in scientific notation calculations, the measurement with the fewest significant figures determines your final answer's precision.

Question 20

A laboratory balance reads 15.47 g, 15.46 g, and 15.48 g for three consecutive measurements of the same object. If the true mass is 15.52 g, which statement best describes these measurements?

  1. The measurements are both accurate and precise with systematic error present
  2. The measurements are precise but not accurate due to systematic error (correct answer)
  3. The measurements are accurate but not precise due to random error
  4. The measurements are neither accurate nor precise due to both errors
  5. The measurements are both accurate and precise with no significant error
Explanation: When you encounter measurement analysis questions, you need to distinguish between two key concepts: accuracy (how close measurements are to the true value) and precision (how close measurements are to each other). Let's analyze these measurements: 15.47 g, 15.46 g, and 15.48 g, with a true mass of 15.52 g. First, check precision by examining the spread of values. The measurements cluster tightly around 15.47 g, with only 0.02 g variation between the highest and lowest readings. This indicates high precision. Next, assess accuracy by comparing the average measurement (15.47 g) to the true value (15.52 g). There's a consistent 0.05 g difference, showing the measurements are systematically low. This systematic deviation from the true value indicates poor accuracy due to systematic error—likely from an improperly calibrated balance. Choice A is incorrect because the measurements aren't accurate—they consistently underestimate the true mass. Choice C is wrong because the measurements are actually quite precise (very similar to each other), and the error is systematic, not random. Random error would cause measurements to scatter both above and below the true value. Choice D fails because the measurements clearly demonstrate good precision. Choice B correctly identifies that the measurements are precise (consistent with each other) but not accurate (systematically different from the true value) due to systematic error. Remember: precision relates to reproducibility among measurements, while accuracy relates to closeness to the true value. Systematic errors affect accuracy but don't necessarily impact precision.