Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

$\prod =MRT$ (1)

Here, $\prod$ is the osmotic pressure.

$M$ is the molarity of the solution.

$R$ is universal gas constant.

$T$ is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (2)

The conversion factor to convert $°\text{C}$ to ${}^{°}\text{K}$ is as follows:

$T\left({}^{°}\text{K}\right)=T\left({}^{°}\text{C}\right)+273.15\text{K}$

Answer to Problem 13.139P

The molar mass of the fragment is $\text{1}\text{.82}\times {10}^4\text{g/mol}$.

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

$M=\frac{\prod }{RT}$ (3)

Substitute $0.340\text{torr}$ for $\prod$, $\text{0}\text{.0821}\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ and ${\text{25}}^{°}\text{C}$ for $T$ in equation (3).

$\begin{array}{c}M=\frac{\left(0.340\text{torr}\right)\left(\frac{1\text{atm}}{760\text{torr}}\right)}{\left(\text{0}\text{.0821}\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left({\text{25}}^{°}\text{C}+\text{273}\text{.15}\right)\text{K}}\\ =1.828544\times {10}^{-5}M\end{array}$

Rearrange equation (2) to calculate the amount of solute as follows:

$\text{Amount}\text{of}\text{solute}=\left(\text{Molarity}\right)\left(\text{Volume of}\text{}\text{solution}\right)$ (4)

Substitute $1.828544\times {10}^{-5}M$ for the molarity and $\text{30}\text{mL}$ for the volume of solution in equation (4).

$\begin{array}{c}\text{Amount}\text{of}\text{solute}=\left(\frac{1.828544\times {10}^{-5}\text{mol}}{1\text{L}}\right)\left(\text{30}\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\\ =5.48563\times {10}^{-5}\text{mol}\end{array}$

The molar mass of the compound is calculated as follows:

$\text{Molar mass}=\frac{\text{Given mass}}{\text{Number of moles}}$ (5)

Substitute $5.48563\times {10}^{-5}\text{mol}$ for the number of moles and $\text{10}\text{mg}$ for the given mass in equation (5).

$\begin{array}{c}\text{Molar mass}=\frac{\left(\text{10}\text{mg}\right)\left(\frac{{10}^{-3}\text{g}}{1\text{mg}}\right)}{5.48563\times {10}^{-5}\text{mol}}\\ =1.82294\times {10}^4\text{g/mol}\\ \text{=1}\text{.82}\times {10}^4\text{g/mol}\end{array}$

Conclusion

The molar mass of the fragment is $\text{1}\text{.82}\times {10}^4\text{g/mol}$.

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

$\Delta T_{\text{f}}=i{k}_{\text{f}}m$ (6)

Here, $\Delta T_{\text{f}}$ is the change in freezing point.

$i$ is van’t Hoff factor.

$k_{\text{f}}$ is the freezing point depression constant.

$m$ is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by $m$ and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

$\text{Density of solution}=\frac{\text{Mass of solution}}{\text{Volume of solution}}$ (7)

Answer to Problem 13.139P

The freezing point of the solution is $-\text{3}\text{.41}\times {10}^{-5}{}^{°}\text{C}$.

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

$\text{Mass of solution}=\left(\text{Density of solution}\right)\left(\text{Volume of solution}\right)$ (8)

Substitute $0.997\text{g/mL}$ for the density of the solution and $\text{30}\text{mL}$ for the volume of solution in equation (8).

$\begin{array}{c}\text{Mass of solution}=\left(\frac{0.997\text{g}}{1\text{mL}}\right)\left(\text{30}\text{mL}\right)\\ =29.91\text{g}\end{array}$

The formula to calculate the mass of the solution is as follows:

$\text{Mass of solution}=\text{Mass of solute}+\text{Mass of solvent}$ (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}$ (10)

Substitute $29.91\text{g}$ for the mass of solution and $\text{10}\text{mg}$ for the mass of solute in equation (10) to calculate the mass of water.

$\begin{array}{c}\text{Mass of water}=\left(29.91\text{g}-\left(\text{10 mg}\right)\left(\frac{{10}^{-3}\text{g}}{1\text{mg}}\right)\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\\ =0.0299\text{kg}\end{array}$

The formula to calculate the molality of the solution is as follows:

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$ (11)

Substitute $5.48563\times {10}^{-5}\text{mol}$ for the number of moles and $0.0299\text{kg}$ for the mass of solvent in equation (11).

$\begin{array}{c}\text{Molality}=\frac{5.48563\times {10}^{-5}\text{mol}}{0.0299\text{kg}}\\ =1.83466\times {10}^{-5}m\end{array}$

Substitute $1.83466\times {10}^{-5}m$ for $m$, 1 for $i$, ${1.86}^{°}\text{C/}m$ for $k_{\text{f}}$ in equation (6).

$\begin{array}{c}\Delta T_{\text{f}}=i{k}_{\text{f}}m\\ =\left(1\right)\left({1.86}^{°}\text{C/}m\right)\left(1.83466\times {10}^{-5}m\right)\\ =3.412\times {10}^{-5}{}^{°}\text{C}\\ =\text{3}\text{.41}\times {10}^{-5}{}^{°}\text{C}\end{array}$

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

$T_{\text{f}}=T_{\text{f}\left(\text{solvent}\right)}-T_{\text{f}\left(\text{solution}\right)}$ (12)

Substitute $0^{°}\text{C}$ for $T_{\text{f}\left(\text{solvent}\right)}$ and $3.412\times {10}^{-5}{}^{°}\text{C}$ for  $T_{\text{f}\left(\text{solution}\right)}$ in equation (120.

$\begin{array}{c}{T}_{\text{f}}=T_{\text{f}\left(\text{solvent}\right)}-T_{\text{f}\left(\text{solution}\right)}\\ =0^{°}\text{C}-3.412\times {10}^{-5}{}^{°}\text{C}\\ =-3.412\times {10}^{-5}{}^{°}\text{C}\\ =-\text{3}\text{.41}\times {10}^{-5}{}^{°}\text{C}\end{array}$

Conclusion

The freezing point of the solution is $-\text{3}\text{.41}\times {10}^{-5}{}^{°}\text{C}$.