AP Chemistry - Other Solution Concepts

A solution on NaCl has a denisty of 1.075 g/mL. If there are 0.475 L of solution present, what is the mass?

510.6 g

441.9 g

0.510 g

2.263 g

2.2 g

Explanation

1.075 g / mL * 0.475 L

First, convert to mL

1.075 g / mL * 475 mL = 510.6 g

A solution was prepared by dissolving 22.0 grams of in water to give a 110mL solution. What is the concentration in 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 to calculate the moles of solute:

$MW_{NaOH}=(23\frac{g}{mole})+(16\frac{g}{mole})+(1\frac{g}{mole})=40\frac{g}{mole}$

$22g*\frac{mole}{40g}=0.55\ moles\ NaOH$

$\frac{0.55\ moles\ NaOH}{0.110L}=5M$

A solution was prepared by dissolving 40.0 grams of $CH_{3}CH_{2}OH$ in water to give a 50mL solution. What is the concentration in 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 $CH_{3}CH_{2}OH$ to calculate the moles of solute:

$MW=(atomic\ weight\ of\ C)+(atomic\ weight\ of\ H)+(atomic\ weight\ of\ O)$

$MW_{CH_{3}CH_{2}OH}=2(12\frac{g}{mole})+6(1g\frac{g}{mole})+(16\frac{g}{mole})=46 \frac{g}{mole}$

$40.0g*\frac{mole}{46g}=0.87\ moles\ CH_{3}CH_{2}OH$

$\frac{0.87\ moles\ CH_{3}CH_{2}OH}{0.050\L}=17M$

How many milliliters of solution is needed to dissolve 5 grams of to prepare a solution of concentration 10M?

Explanation

In order to calculate the number of milliliters, we must first determine the number of moles in 5 grams of using its molecular weight as a conversion factor:

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

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

$5g*\frac{mole}{58g}= 0.086\ moles$

Using the concentration units as a conversion factor and the number of moles calculated, the number of milliliters can be calculated:

$0.086 moles * \frac{L}{10\ moles}= 0.0086\ L = 8.6\ mL$

What volume of water must be added to 750mL of 0.050M sodium chloride () in order to achieve a final concentration of 0.015M?

Explanation

For a solution of known volume and concentration (molarity in this case), the volume needed to dilute the solution to a desired concentration may be found using the formula:

$M_{1}V_{1} = M_{2} V_2$

Where and are the initial and final concentrations, and and are the initial and final volumes. So, for 750mL (0.750L) of a 0.050M solution diluted to 0.015M:

Solving for :

$V_2 = \frac{0.050 M * 0.750L}{0.015 M} = 2.5 L$

Now that we know the total volume needed, we may find the volume that must be added by subtracting the initial volume () from the final volume ():

1.75L of water must be added to 750mL of 0.050M in order to achieve a final concentration of 0.015M

What is the osmotic pressure of a 5.0M solution of $\small C_{2}H_{6}$ at $\small 10^{o}C$ ?

$\small 116 atm$

$\small 1 atm$

$\small 232 atm$

$\small 4 atm$

Explanation

Osmotic pressure is represented by:

$\small \Pi = iMRT$

Where $\small i=$ Van’t hoff factor, $\small M = Molarity$ , $\small R =$ gas constant $\left(0.08206 \frac{L\cdot atm}{mol\cdot K}\right)$ , $\small T =$ temperature in $\small K$ . The Van’t hoff factor is a unitless number that represents the amount of ionic species that the compound $\small (C_{2}H_{6})$ will dissociate in solution. $\small C_{2}H_{6}$ is part of a large group of molecules classified as hydrocarbons which normally do not dissociate at all in solution. Therefore, $\small i = 1$ .

Plug in known values and solve.

$\Pi = 1\cdot 5\cdot 0.08206\frac{L\cdot atm}{mol\cdot K}\cdot 283$

$\Pi= 116atm$

Which of the following is a weak electrolyte?

Explanation

Solutes that dissociate completely in a solution are called strong electrolytes. Weak electrolytes stay paired to some extent in solutions. As a result, strong electrolytes include ionic compounds and strong acid and bases.

Suppose that two containers, and , contain equal amounts of water. If 5 moles of is added to solution and 5 moles of glucose is added to solution , which solution will experience a greater increase in boiling point?

Solution , because is able to dissociate into $Na^{+}$ and $Cl^{-}$ ions, thus resulting in a greater amount of particles dissolved in solution

Solution , because glucose has a greater molar mass than

Neither solution will experience a change in boiling point

Both solutions will exhibit the same change in boiling point

There is not enough information to answer the question

Explanation

In the question stem, we are told that equal molar amounts of and glucose are added to containers and , respectively. The change in boiling point of water is a colligative property that is dependent on the number of dissolved solute particles, regardless of their identity. The addition of 5 moles of will result in approximately 10 moles of dissolved solute, since each mol of can dissociate into two ions, according to the following reaction:

$NaCl(s)\rightarrow Na^{+}\left( aq\right )+Cl^{-}\left( aq\right )$

Glucose, on the other hand, does not dissociate and simply remains as intact molecules. Thus, the addition of 5 moles of glucose to container results in 5 moles of dissolved solute. Since solution contains approximately twice as many dissolved solute particles as does solution , it will experience a greater increase in the boiling point of water.

Which of the following definitions is false?

Solubility product, $K_{sp}$ , is the product of ion concentrations at equilibrium in a supersaturated solution of salt.

Ion-product constant of water, $K_{w}$ , is the product of equilibrium concentration of and ions in an aqueous solution at $25^\circ C$ .

The van't Hoff factor, i, is the number of ions that a compound produces in a solution.

Molality is the number of moles of solute in a solution divided by the number of kilogram of solvent.

Explanation

Solubility product, $K_{sp}$ , is the product of ion concentrations at equilibrium in a saturated solution of salt. All other definitions are true.

A solution was prepared by diluting 10mL of a 0.500M salt solution to 20mL. What would be the final concentration of this solution?