Chemistry - Liquid Solution Question with Solution | TestHub

The problem involves the concept of relative lowering of vapor pressure, a colligative property. Raoult's Law for non-volatile solutes states that the relative lowering of vapor pressure is equal to the mole fraction of the solute.

Raoult's Law:

$$\frac{P^0-P_S}{P^0}=x_{solute}=\frac{n_{solute}}{n_{solute}+n_{solvent}}$$

For dilute solutions, this can be approximated as:

$$\frac{P^0-P_S}{P_S}=\frac{n_{solute}}{n_{solvent}}$$

Where:

$P^0$ = vapor pressure of pure solvent

$P_S$ = vapor pressure of solution

$n_{solute}$ = moles of solute

$n_{solvent}$ = moles of solvent

Given:

Mass of solute = 8 gm

Mass of water (solvent) = 200 gm

Molar mass of water ($M_{H_{2}O}$) = 18 gm/mol

Initial vapor pressure of solution ($P_{S1}$) = 25 torr

2 moles of water are added, so new mass of water = $200+(2\times 18)=200+36=236$ gm

New vapor pressure of solution ($P_{S2}$) = $25+0.5=25.5$ torr

Let $M$ be the molar mass of the non-volatile solute.

Moles of solute ($n_{solute}$) = $\frac{8}{M}$

Initial moles of water ($n_{solvent1}$) = $\frac{200}{18}$

Final moles of water ($n_{solvent2}$) = $\frac{236}{18}$

Using the approximate form of Raoult's Law:

Case 1: Initial solution

$$\frac{P^0-25}{25}=\frac{8/M}{200/18}=\frac{8}{M}\times \frac{18}{200}\dots (1)$$

Case 2: After adding 2 moles of water

$$\frac{P^0-25.5}{25.5}=\frac{8/M}{236/18}=\frac{8}{M}\times \frac{18}{236}\dots (2)$$

From (1), we can express $\frac{8}{M}\times 18$:

$$\frac{8}{M}\times 18=\frac{P^0-25}{25}\times 200=8(P^0-25)\dots (3)$$

Substitute (3) into (2):

$$\frac{P^0-25.5}{25.5}=\frac{8(P^0-25)}{236}$$

$$236(P^0-25.5)=25.5\times 8(P^0-25)$$

$$236P^0-236\times 25.5=204P^0-204\times 25$$

$$236P^0-6018=204P^0-5100$$

$$236P^0-204P^0=6018-5100$$

$$32P^0=918$$

$$P^0=\frac{918}{32}$$

$$P^0=28.6875torr$$

Rounding to two decimal places, the vapor pressure of pure water is 28.69 torr.

The final answer is $\boxed{28.68torr}$.