Identifying Unknown Concentration - AP Chemistry

Card 0 of 60

Question

What is the concentration if 40 g NaOH is dissolved in 1000 g of water

Answer

first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent

you have 40 g NaOH * 1 mol/40 g = 1 mol

1000 g of water is equivalent to 1 L

1 mol/L = 1M

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Question

80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?

Answer

Molarity = mol solute / L soution

mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol

L solution = 50000 mL water * 1 L/1000 mL = 50 L

2 mol / 50 L =

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Question

What is the molality of a solution created by mixing 4.3g $NaCl$ into 43g of water?

Answer

Molality can be defined:

$Molality=\hspace{1 mm}\frac{mol\hspace{1 mm}solute}{kg\hspace{1 mm}solvent}$

It is slightly different from Molarity and has different uses.

$Molality=\hspace{1 mm}\frac{4.3\hspace{1 mm}g\hspace{1 mm}NaCl}{43\hspace{1 mm}g\hspace{1 mm}H_20}\times\frac{1000\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}kg\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaCl}{58.44\hspace{1 mm}g\hspace{1 mm}NaCl}=1.71\hspace{1 mm}m$

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Question

How much solid NaOH must be dissolved to make 740mL of a 0.32M solution?

Answer

This problem can be solved by stoichiometry. Remember that 0.32M gives us the moles of NaOH per liter, and solve for the number of moles per 0.740L.

$740\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.32\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=9.47\hspace{1 mm}g\hspace{1 mm}NaOH$

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Question

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

$\rho_{ethanol}=0.789\frac{g}{mL}$

Answer

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

$25\hspace{ 1 mm}mL\hspace{ 1 mm}EtOH\times\frac{0.789\hspace{ 1 mm}g}{1\hspace{ 1 mm}mL}=19.7\hspace{ 1 mm}g\hspace{ 1 mm}EtOH$

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

$19.7\hspace{ 1 mm}g+75.0\hspace{ 1 mm}g= 94.7\hspace{ 1 mm}g$

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Question

How many liters of 10M HCl are needed to make 4L of 0.4M solution?

Answer

A simple calculation can be done to perform any solution dilution problem. We know our equation $molarity = \frac{moles}{volume}$ .

We can rewrite this as .

Using this formula, we take the old solution and set it equal to the new solution.

$V = \frac{1.6mol}{10M}=0.16L$

We need 0.16 liters of our 10 molar solution.

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Question

Find the mass of in 10L of water if it is a 2m solution.

Answer

Molality is grams of solute per kilogram of solvent.

$m=\frac{mol}{kg}\rightarrow mol=(m)(kg)$

Water has a density of one gram per mililiter, so one liter of water equal to one kilogram. If we have a 2m solution, that means we have two moles of per kilogram of water.

$10L\frac{1kg}{1L}=10kg$

has a molecular weight of $\small 56\frac{g}{mol }$ .

$20mol\frac{56g}{1mol}=1120g\ KOH$

This gives us of .

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Question

Which solution has the highest molarity?

Answer

This question requires us to calculate molarity for each answer choice. It is important to add everything correctly and be careful with more complex compounds.

Molarity is simply moles of solute over liters of solution. The correct answer, after trying each, is the answer with lead (II) nitrate, as it gets us a molarity of 2.

$Pb(NO_3)_2:\ \frac{486.5g\frac{1mol}{331g}}{0.75L}=1.96M$

$NaCl:\ \frac{174g\frac{1mol}{58.5g}}{2.0L}=1.49M$

$MgSO_4:\ \frac{240g\frac{1mol}{120.3g}}{3.0L}=0.67M$

$NaNO_3:\ \frac{85g\frac{1mol}{85g}}{2.0L}=0.50M$

$CH_3CH_2OH:\ \frac{138g*\frac{1mol}{46g}}{3.0L}=1.0M$

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Question

Which of the following aqueous solutions is the most concentrated?

$\small \rho_{water}=1 \frac{g}{mL}$

Answer

In order to answer this question, it helps to know that 1 kilogram of water is equal to 1 liter of water, due to its density. Two of the above options refer to a 1m solution of hydrochloric acid. The other is a 1M solution.

$1M\ HCl=\frac{1mol\ HCl}{1L}$

$1m\ HCl=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

$36.5g\ \text{HCl in}\ 1kg\ \text{water}=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

All three of the options have the same amount of hydrochloric acid (one mole). For molarity, the hydrochloric acid is diluted with water until one liter of solution is created. For molality, one mole of HCl is added to one kilogram of water. Since one kilogram of water is one liter, this becomes the same concentration.

One a very small level, the 1M HCl solution will be slightly more concentrated. Creating a molal solution does not take into account the volume of the solute. If, for example, 100 cubic centimeters of HCl were added to one kilogram of water, the resulting volume would be more than one liter, making the concentration slightly less than 1M. This discrepancy is usually not accounted for in basic chemistry, but you should be familiar with the concept.

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Question

10mL of a solution of of unknown concentration mixed with 34mL of $10^{-4}M$ produced a solution with a pH of 7. What is the concentration of the solution?

Answer

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=3.4\cdot 10^{-4}M$

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Question

30mL of a solution of of unknown concentration mixed with 19mL of $10^{-3}M$ produced a solution with a pH of 7. What is the concentration of the solution?

Answer

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=6.3\cdot10^{-4}M$

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Question

39mL of a solution of $10^{-4}M$ is mixed with 24mL of an solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the solution?

Answer

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

$\frac{N_B M_B V_B }{N_A V_A}= M_A$

Plug in known values and solve.

$M_A=1.63\cdot10^{-4}M$

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Question

29mL of a solution of $10^{-4}M$ is mixed with 65mL of an solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the solution?

Answer

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

$\frac{N_B M_B V_B }{N_A V_A}= M_A$

Plug in known values and solve.

$M_A=4.5\cdot10^{-4}M$

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Question

It takes 25mL of a .001M to neutralize 5mL of a solution of of unknown concentration. What is the concentration of the unknown solution?

Answer

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

$\frac{N_B M_B V_B }{N_A V_A}= M_A$

Plug in known values and solve.

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Question

It takes 35mL of .01M to neutralize 5mL of a solution of of unknown concentration. What is the concentration of the unknown solution?

Answer

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

Plug in known values and solve.

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Question

What is the concentration if 40 g NaOH is dissolved in 1000 g of water

Answer

first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent

you have 40 g NaOH * 1 mol/40 g = 1 mol

1000 g of water is equivalent to 1 L

1 mol/L = 1M

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Question

80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?

Answer

Molarity = mol solute / L soution

mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol

L solution = 50000 mL water * 1 L/1000 mL = 50 L

2 mol / 50 L =

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Question

What is the molality of a solution created by mixing 4.3g $NaCl$ into 43g of water?

Answer

Molality can be defined:

$Molality=\hspace{1 mm}\frac{mol\hspace{1 mm}solute}{kg\hspace{1 mm}solvent}$

It is slightly different from Molarity and has different uses.

$Molality=\hspace{1 mm}\frac{4.3\hspace{1 mm}g\hspace{1 mm}NaCl}{43\hspace{1 mm}g\hspace{1 mm}H_20}\times\frac{1000\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}kg\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaCl}{58.44\hspace{1 mm}g\hspace{1 mm}NaCl}=1.71\hspace{1 mm}m$

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Question

How much solid NaOH must be dissolved to make 740mL of a 0.32M solution?

Answer

This problem can be solved by stoichiometry. Remember that 0.32M gives us the moles of NaOH per liter, and solve for the number of moles per 0.740L.

$740\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.32\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=9.47\hspace{1 mm}g\hspace{1 mm}NaOH$

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Question

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

$\rho_{ethanol}=0.789\frac{g}{mL}$

Answer

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

$25\hspace{ 1 mm}mL\hspace{ 1 mm}EtOH\times\frac{0.789\hspace{ 1 mm}g}{1\hspace{ 1 mm}mL}=19.7\hspace{ 1 mm}g\hspace{ 1 mm}EtOH$

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

$19.7\hspace{ 1 mm}g+75.0\hspace{ 1 mm}g= 94.7\hspace{ 1 mm}g$

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