Question

If the mass of absorbent increased by 0.24 gm and the mass of pure solvent (water) decreased by 0.02 gm, then the mass percent of solute (glucose) in its aqueoue solution is:

• A. $\frac{1000}{21}\%$
• B. $\frac{100}{12}\%$
• C. $\frac{1000}{22}\%$
• D. $\frac{100}{11}\%$

Answer 1

Answer: (A)

$\frac{1000}{21}\%$

Answer 2

In the Ostwald-Walker method, dry air is passed through the solution and then through the pure solvent. The decrease in mass of the solution vessels ($w_s$) is proportional to the vapor pressure of the solution ($P_s$), and the decrease in mass of the pure solvent vessels ($w_p$) is proportional to the lowering of vapor pressure ($P_0 - P_s$). The increase in mass of the absorbent is the total mass of solvent vapor absorbed, which is $w_s + w_p$.

According to the method, $w_s \propto P_s$ and $w_p \propto (P_0 - P_s)$. Thus, $\frac{w_p}{w_s} = \frac{P_0 - P_s}{P_s}$.

From Raoult's law for a dilute solution of a non-volatile solute, $\frac{P_0 - P_s}{P_s} = \frac{n_{solute}}{n_{solvent}}$, where $n_{solute}$ and $n_{solvent}$ are the number of moles of solute and solvent, respectively.

Given: Increase in mass of absorbent = 0.24 gm. This is $w_s + w_p$. Decrease in mass of pure solvent = 0.02 gm. This is $w_p$. So, $w_s + w_p = 0.24$ gm and $w_p = 0.02$ gm. The decrease in mass of the solution is $w_s = (w_s + w_p) - w_p = 0.24 - 0.02 = 0.22$ gm.

Now, we have $\frac{w_p}{w_s} = \frac{0.02}{0.22} = \frac{2}{22} = \frac{1}{11}$. Using the relationship from Ostwald-Walker method and Raoult's law: $\frac{n_{solute}}{n_{solvent}} = \frac{w_p}{w_s} = \frac{1}{11}$.

The solute is glucose (molar mass $M_{solute} = 180$ g/mol) and the solvent is water (molar mass $M_{solvent} = 18$ g/mol). Let $m_{solute}$ be the mass of glucose and $m_{solvent}$ be the mass of water in the solution. $n_{solute} = \frac{m_{solute}}{M_{solute}} = \frac{m_{solute}}{180}$. $n_{solvent} = \frac{m_{solvent}}{M_{solvent}} = \frac{m_{solvent}}{18}$.

So, $\frac{n_{solute}}{n_{solvent}} = \frac{m_{solute}/180}{m_{solvent}/18} = \frac{m_{solute}}{m_{solvent}} \times \frac{18}{180} = \frac{m_{solute}}{m_{solvent}} \times \frac{1}{10}$.

We have $\frac{n_{solute}}{n_{solvent}} = \frac{1}{11}$. So, $\frac{m_{solute}}{m_{solvent}} \times \frac{1}{10} = \frac{1}{11}$. $\frac{m_{solute}}{m_{solvent}} = \frac{10}{11}$.

We need to find the mass percent of solute in the aqueous solution. Mass percent of solute = $\frac{m_{solute}}{m_{solute} + m_{solvent}} \times 100\%$. Let $m_{solute} = 10x$ and $m_{solvent} = 11x$. Mass percent = $\frac{10x}{10x + 11x} \times 100\% = \frac{10x}{21x} \times 100\% = \frac{10}{21} \times 100\% = \frac{1000}{21}\%$.