The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?

a. 16mm Hg

b. 24mm Hg

c. 20mm Hg

d. 12mm Hg

Answer: The vapor pressure (VP) of this solution at $25^{o}C$ is closest to the value 24 mm Hg.

Explanation:

Given: Mass of sucrose = 171 g

Mass of water = 1 L = 1000 g

Vapor pressure of water = 24 mm Hg

As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{1000 g}{18.02 g/mol}\\= 55.49 mol$

Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{171 g}{342 g/mol}\\= 0.5 mol$

Total moles = 55.49 + 0.5 mol = 55.99 mol

Mole fraction of water is as follows.

$Mole fraction = \frac{moles of water}{total moles}\\= \frac{55.49}{55.99}\\= 0.99$

Formula used to calculate vapor pressure of the solution is as follows.

$P_{i} = P^{o}_{i} \times \chi_{i}$

where,

$P_{i}$ = vapor pressure of component i over the solution

$P^{o}_{i}$ = vapor pressure of pure component i

$\chi_{i}$ = mole fraction of i

Substitute the values into above formula to calculate vapor pressure of water as follows.

$P_{i} = P^{o}_{i} \times \chi_{i}\\= 24 mm Hg \times 0.99\\= 23.76 \\or 24 mm Hg\\$

Thus, we can conclude that the vapor pressure (VP) of this solution at $25^{o}C$ is closest to the value 24 mm Hg.

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