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Step-by-Step Solution
Step 1: Identify the Known Data
• Vapour pressure of pure water at 20°C, $P^\circ = 17.5 \text{ mm Hg}$.
• Mass of glucose (solute), $m_{\text{glucose}} = 18 \text{ g}$.
• Molar mass of glucose, $M_{\text{glucose}} = 180 \text{ g mol}^{-1}$.
• Mass of water (solvent), $m_{\text{water}} = 178.2 \text{ g}$.
• Molar mass of water, $M_{\text{water}} = 18 \text{ g mol}^{-1}$.
Step 2: Calculate the Moles of Glucose and Water
• Moles of glucose,
$$
n_{\text{glucose}} = \frac{m_{\text{glucose}}}{M_{\text{glucose}}}
= \frac{18}{180}
= 0.1 \text{ mol}.
$$
• Moles of water,
$$
n_{\text{water}} = \frac{m_{\text{water}}}{M_{\text{water}}}
= \frac{178.2}{18}
= 9.9 \text{ mol (approximately)}.
$$
Step 3: Write the Vapour Pressure Relation
According to Raoult’s law for a non-volatile solute (glucose), the lowering of vapour pressure can be expressed as:
$$
\frac{P^\circ - P_S}{P_S} = \frac{n_{\text{glucose}}}{n_{\text{water}}}.
$$
Here:
• $P^\circ$ = vapour pressure of pure solvent (water) = 17.5 mm Hg
• $P_S$ = vapour pressure of solution
• $n_{\text{glucose}}$ = moles of glucose
• $n_{\text{water}}$ = moles of water
Step 4: Substitute the Known Values
Substituting $P^\circ = 17.5$, $n_{\text{glucose}} = 0.1$, and $n_{\text{water}} = 9.9$ into the equation:
$$
\frac{17.5 - P_S}{P_S} = \frac{0.1}{9.9}.
$$
Step 5: Solve for $P_S$
First, compute the right-hand side:
$$
\frac{0.1}{9.9} = \frac{1}{99} \approx 0.0101.
$$
Thus the equation becomes:
$$
\frac{17.5 - P_S}{P_S} = 0.0101.
$$
Rearrange and solve step by step:
(a) Multiply both sides by $P_S$:
$$
17.5 - P_S = 0.0101 \times P_S.
$$
(b) Bring like terms together:
$$
17.5 = P_S + 0.0101 \, P_S = P_S(1 + 0.0101).
$$
(c) Factor out $P_S$:
$$
17.5 = 1.0101 \, P_S.
$$
(d) Finally, solve for $P_S$:
$$
P_S = \frac{17.5}{1.0101} \approx 17.325 \text{ mm Hg}.
$$
Step 6: State the Final Answer
Therefore, the vapour pressure of the solution at 20°C is
$$
P_S = 17.325 \text{ mm Hg}.
$$
Hence, option (4) is the correct answer.
