Step 1: Understanding the Question:
Relative lowering of vapour pressure (RLVP) is a colligative property equal to the mole fraction of the solute in the solution.
Step 2: Key Formula or Approach:
\[ \text{RLVP} = \frac{\text{P}^\circ - \text{P}}{\text{P}^\circ} = \text{X}_{\text{solute}} = \frac{n_2}{n_1 + n_2} \]
For a very dilute solution, \( n_1 + n_2 \approx n_1 \). Thus, \( \text{RLVP} \approx \frac{n_2}{n_1} \).
Step 3: Detailed Explanation:
1. Moles of solute (\( n_2 \)):
\[ n_2 = \frac{\text{Mass of solute}}{\text{Molar mass}} = \frac{0.56}{60} \approx 0.00933 \text{ mol} \]
2. Moles of solvent (\( n_1 \), water):
\[ n_1 = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{100}{18} \approx 5.556 \text{ mol} \]
3. Calculate RLVP:
\[ \text{RLVP} \approx \frac{0.00933}{5.556} \approx 0.001679 \]
Rounding to four decimal places, we get 0.0017.
Step 4: Final Answer:
The relative lowering of vapour pressure is 0.0017.