Question

Calculate the relative lowering of vapor pressure of a solution containing 0.56 g nonvolatile solute in 100 g water [molar mass of solute = 60 g mol\(^{-1}\)].

• A. \( 0.0024 \)
• B. \( 0.0120 \)
• C. \( 0.0017 \)
• D. \( 0.0221 \)

Hint:

The relative lowering of vapor pressure is determined by the mole fraction of the solute, which depends on the ratio of moles of solute to solvent.

Answer 1

The Correct Option is A

Step 1: Use Raoult's Law.
The relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P_0} = \frac{\text{mol of solute}}{\text{mol of solvent}} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]
where \( P_0 \) is the vapor pressure of the pure solvent, and \( \Delta P \) is the lowering of vapor pressure.

Step 2: Calculate moles of solute and solvent.
The number of moles of the solute is:
\[ n_{\text{solute}} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{0.56}{60} = 0.00933 \, \text{mol} \]
The number of moles of solvent (water) is:
\[ n_{\text{solvent}} = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{100}{18} = 5.56 \, \text{mol} \]

Step 3: Calculate relative lowering of vapor pressure.
Now, calculate the relative lowering of vapor pressure:
\[ \frac{\Delta P}{P_0} = \frac{0.00933}{5.56} = 0.0024 \]

Step 4: Final conclusion.
Thus, the relative lowering of vapor pressure is:
\[ \boxed{0.0024} \]