Concentration and Units - AP Chemistry

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 =

The definition of molality is moles of solute in 1 kg of the solvent, whereas molarity is the number of moles of solute per 1 L of solutioin. Since 1 mol of CaCl2 is added to 1 L of water, this means that the volume of the final solution is greater than 1 L. Thus, molality is the more accurate concentration determinant, since the solution is probably close to 1 L.

Molarity, molality, and normality are the three principle ways to measure concentration. Molarity is a measure of moles of solute per liter of solution. Molality is a measure of moles of solute per kilogram of solvent. Normality expressly relates to acids and bases, and is the measure of moles of solute divided by the number of hydrogen equivalents per mole, all divided by liters of solution. Normality is also referred to as "equivalents (of acid) per liter."

Use the equation M1V1 = M2V2, and plug in the corresponding values to solve for V2:

V2 = (3)(50)/(0.75) = 200mL

We started out 50mL of the solution, so you have to add 150mL to get a final volume of 200mL.

Molality can be defined:

Molality=hspace{1 mm}$\frac{molhspace{1 mm}$solute}{kghspace{1 mm}solvent}

It is slightly different from Molarity and has different uses.

Molality=hspace{1 mm}$\frac{4.3hspace{1 mm}$ghspace{1 mm}NaCl}{43hspace{1 mm}ghspace{1 mm}H_20}times$\frac{1000hspace{1 mm}$ghspace{1 mm}H_2O}{1hspace{1 mm}kghspace{1 mm}H_2O}times$\frac{1hspace{1 mm}$molehspace{1 mm}NaCl}{58.44hspace{1 mm}ghspace{1 mm}NaCl}=1.71hspace{1 mm}m

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.

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.

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.

A simple calculation can be done to perform any solution dilution problem. We know our equation .

We can rewrite this as .

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

We need 0.16 liters of our 10 molar solution.

Molality is grams of solute per kilogram of solvent.

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.

has a molecular weight of .

This gives us of .

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.

In order to dilute the concentrated acid, we need to find the amount of concentrated acid that will be diluted to 50mL of total solution. We can find the volume of concentrated acid necessary by setting the final volume and concentration equal to the initial concentration and unknown volume.

The initial concentration is 2M, the final concentration is 0.5M, and the final volume is 50mL

This means that 12.5mL of concentrated acid needs to be diluted to 50mL of solution. This will result in a solution with a concentration of 0.5M.

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.

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.

To solve this problem, we may use the following equation relating the molarity and volume of two solutions:

Recall:

Plug in known values and solve.

However, this is not the final answer. The whole volume of the second, 0.4M solution is 85mL. Thus the chemist needs to add 25mL of water to the original solution to obtain the desired concentration.

Use the dilution formula:

Rearranging this equation gives:

Plugging in the values gives:

Use the dilution formula:

Rearranging this equation gives:

Plugging in the values gives:

Therefore, after diluting the solution to 50mL, the solution concentration would be lowered from 0.35M to 0.21M.

By using the concentration as a conversion factor, the number of moles can calculated by multiplying the concentration by the number of liters.

Therefore, there are in of a solution.

In order to calculate the concentration, we must use molarity formula:

We must use the molecular weight of sodium chloride to calculate the moles of solute:

Therefore, the concentration in molarity of this solution is 0.040M.

By simply using the concentration as a conversion factor, the number of moles can be calculated by multiplying the concentration by the number of liters. Before calculating the number of moles, the number of milliliters must be converted to liters using the fact that .

Therefore, there are in of a solution.

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 =