Q131CP - Chemistry: Molecular Nature Of Matter And Change

Question

A solution of $\mathbf{\text{1.50 g}}$ of solute dissolved in $\mathbf{\text{25.0 mL}}$ of ${\mathbf{H}}_2\mathbf{O}$ at ${\mathbf{25}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$ has a boiling point of $\mathbf{100}\mathbf{.}{\mathbf{45}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$.

(a) What is the molar mass of the solute if it is a non-volatile nonelectrolyte and the solution behaves ideally ($\mathit{d}$ of ${\mathbf{H}}_2\mathbf{O}$ at $\mathbf{\text{25°C}}\mathbf{=}\mathbf{\text{0.997 }}\raisebox{1ex}{$\mathbf{\text{g}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{mL}}$}\right.$)?

(b) Conductivity measurements show the solute to be ionic of general formula ${\mathbf{\text{AB}}}_{\mathbf{\text{2}}}$ or ${\mathbf{\text{A}}}_{\mathbf{\text{2}}}\mathbf{\text{B}}$. What is the molar mass if the solution behaves ideally?

(c) Analysis indicates an empirical formula of ${\mathbf{\text{CaN}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{6}}}$. Explain the difference between the actual formula mass and that calculated from the boiling point elevation.

(d) Find the van’t Hoff factor $\left(i\right)$ for this solution.

Step-by-Step Solution

Answer

(a) If solute is non-volatile nonelectrolyte, the molar mass of solute is $68.56\raisebox{1ex}{${\displaystyle \text{ g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

(b) Molar mass of solute having general ionic formula ${\text{AB}}_{\text{2}}$ or ${\text{A}}_{\text{2}}\text{B}$ is $205.761\text{ }\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

(c) The actual formula mass is less than the calculated mass from boiling point elevation.

(d) The van’t Hoff factor is $3$.

A concept:

The temperature difference between the solution's boiling point and the pure solvent's boiling point is known as the boiling point elevation.

 

Consider the given data below.

The mass of solute, $m=1.50\text{ g}$ 

The volume of the solvent, $V=25\text{ mL}$ 

The density of solvent, $\rho =0.997\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.$ 

(a) Molar mass of solute:

To determine the molecular mass of solute which is a non-volatile nonelectrolyte, first, calculate the molality of the solution.

Therefore, the mass of the solvent can be calculated as follow.

$\begin{array}{rcl}M& =& \rho V\\ & =& 0.997\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.\times 25\text{ mL}\\ & =& 24.925\text{ g}\\ & =& 0.0249\text{ kg}\end{array}$

The change in boiling point for a one-mole solution of a nonvolatile molecular solute is the same as the molal boiling-point elevation constant.

 

It can be calculated by following the formula

$T_{\text{b}\left(\text{solution}\right)}-T_{\text{b}\left(\text{solvent}\right)}=\text{molality}\times K_b$                                                                                    ….. (1)

Here, $K_b$ is the molal boiling point elevation constant.

Given that,

The temperature of the solution is,

 $T_{\text{b}\left(\text{solution}\right)}={100.45}^{\text{o}}\text{C}$

The temperature of the solvent is,

$T_{\text{b}\left(\text{solvent}\right)}={100}^{\text{o}}\text{C}$  

The molal boiling point elevation is,

$K_b={0.512}^{\text{o}}\text{C}$  

 

Therefore, by rearranging equation (1) you get

$\begin{array}{rcl}\text{molality}& =& \frac{T_{\text{b}\left(\text{solution}\right)}-T_{\text{b}\left(\text{solvent}\right)}}{K_b}\\ & =& \frac{{(100.45-100)}^{\text{o}}\text{C}}{{0.512}^{\text{o}}\text{C}}\\ & =& 0.8789\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\end{array}$

Molality is defined as the moles of solute dissolved in kilogram of solvent. Hence,

$\text{molality}=\frac{\text{moles\hspace{0.17em}of\hspace{0.17em}solute}}{\text{mass\hspace{0.17em}of\hspace{0.17em}solvent }\left(\text{kg}\right)}$ 

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}solute}& =& \text{molality}\times \text{mass\hspace{0.17em}of\hspace{0.17em}solvent }\left(\text{kg}\right)\\ & =& 0.8789\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\times 0.0249 \text{ kg}\\ & =& 0.02188 \text{ mol}\end{array}$ 

 

Thus, the molar mass of solute can be calculated as;

$\begin{array}{rcl}\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}solute}& =& \frac{\text{mass\hspace{0.17em}of\hspace{0.17em}solute}}{\text{moles\hspace{0.17em}ofsolute}}\\ & =& \frac{\text{1.50 g}}{\text{0.02188 mol}}\\ & =& 68.56\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

Hence, if solute is non-volatile nonelectrolyte, the molar mass of solute is $68.56\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

(b) Molar mass if solution behaves ideally:

As given, ${\text{AB}}_{\text{2}}$ or ${\text{A}}_{\text{2}}\text{B}$ compound will dissociate to give three ions. Thus, the van’t Hoff factor will be $3$.

So, molality would be $\frac{1}{3}$ of $0.8789\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.$.

$\begin{array}{rcl}m& =& \frac{1}{3}\times 0.8789\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\\ & =& 0.2929\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\end{array}$ 

Calculate the moles of solute.

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}solute}& =& m\times \text{mass\hspace{0.17em}of\hspace{0.17em}solvent }\left(\text{kg}\right)\\ & =& 0.2929\text{ mol}/\text{kg}\times 0.0249\text{ kg}\\ & =& 0.00729\text{ mol}\end{array}$ 

 

Therefore, molar mass of solute can be calculated as follow.

$\begin{array}{rcl}\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}solute}& =& \frac{\text{mass\hspace{0.17em}of solute}}{\text{moles\hspace{0.17em}of solute}}\\ & =& \frac{1.50\text{ g}}{0.00729\text{ mol}}\\ & =& 205.761\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

Hence, the molar mass of solute having general ionic formula ${\text{AB}}_{\text{2}}$ or ${\text{A}}_{\text{2}}\text{B}$ is $205.761\text{ }\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

(c) Actual formula mass and that calculated from the boiling point elevation:

Analysis indicates the empirical formula of ${\text{CaN}}_{\text{2}}{\text{O}}_{\text{6}}$. 

Mass of empirical formula is,

$40\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+(2\times 14)\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+(6\times 16)\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.=164\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

The calculated mass from boiling point elevation is $205.761\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

Hence, it is clear that the actual formula mass $\left(164\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)$ is less than the mass that is calculated from the boiling point elevation $\left(205.761\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)$.

(d) Van’t Hoff factor ( i ) :

The van't Hoff factor is the difference between the concentration of a material determined by its mass and the concentration of particles actually formed when the substance is dissolved.

So, the actual concentration or molarity of particles when the substance is dissolved is $0.8789\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.$ and the concentration of substance calculated from its mass is $0.2929\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.$.

Hence, calculate the van’t Hoff factor $\left(i\right)$ as follows;

$\begin{array}{rcl}\left(i\right)& =& \frac{0.8789{\displaystyle \raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}}{0.2929{\displaystyle \raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}}\\ & =& 3\end{array}$ 

Hence, the van’t Hoff factor is $3$.