AP Chemistry - Concentration and Units

What is the concentration of Ca in a solution of 1 mol CaCl2 in 1 L of distilled water? (M = molarity, m= molality)

1 M

2 M

1 m

2 m

Cannot be determined

Explanation

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.

Given that 50 mL of 3M HI was present, what volume would be needed to change the concentration to 0.75M?

150mL

200mL

180mL

50mL

Explanation

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.

In order to dilute a 1mL solution that is 0.01M so that the solution is diluted to $5*10^{-3} M,$ , how many milliliters does this solution need to be diluted to?

Explanation

Use the dilution formula:

$M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\frac{M_{1}V_{1}}{M_{2}}=V_{2}$

Plugging in the values gives:

$V_{2}=\frac{0.01\ M * 1\ mL}{5*10^{-3}\ M}= 2\ mL$

What concentration would you have prepared if you diluted 30mL of a 0.350M salt solution to 50mL?

Explanation

Use the dilution formula:

$M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\frac{M_{1}V_{1}}{V_{2}}=M_{2}$

Plugging in the values gives:

$M_{2}=\frac{0.35\ M * 30\ mL}{50\ mL}= 0.21 M$

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

How many moles are in a 0.010L solution with a concentration that is $5*10^{-3}M$ ?

$5*10^{-5}mol$

$2*10^{-5}mol$

$3*10^{-2}mol$

Explanation

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

$\frac{5*10^{-3}moles}{L} * 0.010L = 5*10^{-5}moles$

Therefore, there are $5*10^{-5}moles$ in of a $5*10^{-3}M$ solution.

In order to prepare a needed solution for an experiment, 0.082 grams of was dissolved in water to give a 35mL solution. What is the molarity of this solution?

Explanation

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

$Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$

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

$MW=(atomic\ weight\ of\ Na)+(atomic\ weight\ of\ Cl)$

$MW_{NaCl}=(23 \frac{g}{mole})+(35 \frac{g}{mole})= 58 \frac{g}{mole}$

$0.082g*\frac{mole}{58g}=0.0014\ moles\ NaCl$

$\frac{0.0014\ moles\ CH_{3}CH_{2}OH}{0.035\L} = 0.040\ M$

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

How many moles are in 1000mL solution with a concentration that is $3*10^{-5}M$ ?

$3*10^{-5}mol$

$8*10^{-5}mol$

$3*10^{-8}mol$

$6.5*10^{-5}mol$

$9*10^{-5}mol$

Explanation

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 $1\ liter = 1000\ mL$ .

$\frac{3*10^{-5}\ moles}{L} * 1\ L = 3*10^{-5}\ moles$

Therefore, there are $3*10^{-5}\ moles$ in $0.010\ L$ of a $5*10^{-3}M$ solution.

A chemist wants to turn a 50.0mL solution of $0.60 M\ HNO_{3}$ into a solution. How much water should she add?

Explanation

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

Recall:

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

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.

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

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

Explanation

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.

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