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.