In the question stem, we are told that equal molar amounts of $\displaystyle NaCl$ and glucose are added to containers $\displaystyle A$ and $\displaystyle B$, 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 $\displaystyle NaCl$ will result in approximately 10 moles of dissolved solute, since each mol of $\displaystyle NaCl$ can dissociate into two ions, according to the following reaction:

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

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

We are given the concentration of $\displaystyle NaOH$ in solution and asked to find the pH. To do this, we must make use of the following equation:

$\displaystyle pOH=-log\left [ OH^{-}\right ]$

It is also important to realize that $\displaystyle NaOH$ is a strong base and will thus dissociate completey according to the following reaction:

$\displaystyle NaOH_{(s)} \rightarrow Na^{+}_{(aq)}+OH^{-}_{(aq)}$

Thus, for every one mol of $\displaystyle NaOH$ that reacts, an equal number of moles of $\displaystyle OH^{-}$ will be produced. And since there are $\displaystyle 3.7\cdot 10^{-4}moles$ $\displaystyle NaOH$ to begin with, then $\displaystyle 3.7\cdot 10^{-4}moles$ $\displaystyle OH^{-}$ will be produced.

$\displaystyle pOH=log(3.7\cdot 10^{-4}M)=3.43$

Remember that this calculated value so far is the pOH, not the pH! To calculate the pH, it is vital to remember that $\displaystyle pH+pOH=14$ at $\displaystyle 25^{o}C$. Thus,

$\displaystyle pH=14-3.43=10.57$

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:

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

Where $\displaystyle M_1$ and $\displaystyle M_2$ are the initial and final concentrations, and $\displaystyle V_1$ and $\displaystyle V_2$ are the initial and final volumes. So, for 750mL (0.750L) of a 0.050M solution diluted to 0.015M:

$\displaystyle 0.050 M * 0.750 L = 0.015 M * V_2$

Solving for $\displaystyle V_2$:

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

Now that we know the total volume needed, we may find the volume that must be added by subtracting the initial volume ($\displaystyle V_1$) from the final volume ($\displaystyle V_2$):

$\displaystyle V_2 - V_1 = 2.5 L - 0.750 L = 1.75 L$

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

Since $\displaystyle HCl$ is a strong acid, calculations should be carried out assuming that the compound dissociates completely:

$\displaystyle HCl\rightarrow H^{+} + Cl^{-}$

$\displaystyle H^+$ and $\displaystyle Cl^-$ are produced in a 1:1 ratio to total dissolved $\displaystyle HCl$, so the concentration of $\displaystyle HCl$ in solution is the same as the concentration of $\displaystyle H^+$:

$\displaystyle \left [ HCl\right ] = \left [ H^{+}\right ] = \left [ Cl^{-}\right ]$

pH is related to the concentration of $\displaystyle H^+$:

$\displaystyle pH = -log[H^{+}]$

$\displaystyle pH = -log[0.025] = 1.60$

Osmotic pressure is represented by:

$\displaystyle \small \Pi = iMRT$

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

Plug in known values and solve.

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

$\displaystyle \Pi= 116atm$

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

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

We must use the molecular weight of $\displaystyle NaOH$ to calculate the moles of solute:

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

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

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

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

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

We must use the molecular weight of $\displaystyle CH_{3}CH_{2}OH$ to calculate the moles of solute:

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

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

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

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

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

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

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

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

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

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

Use the dilution formula:

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

Rearranging this equation gives:

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

Plugging in the values gives:

$\displaystyle M_{2}=\frac{0.500M * 10mL}{20mL}= 0.25M$

Therefore, after diluting the solution to 20mL, the solution concentration would be lowered from 0.50M to 0.25M.

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.