a
According to Raoult's Law

\(\frac{P^{\circ}-P_{s}}{P_{s}}=\frac{W_{B} \times M_{A}}{M_{B} \times W_{A}} \quad \ldots(i)\)

Here \(P^{\circ}=\) Vapour pressure of pure solvent,

\(P_{s}=\) Vapour pressure of solution

\(W_{B}=\) Mass of solute,

\(W_{A}=\) Mass of solvent

\(M_{B}=\) Molar mass of solute,

\(M_{A}=\) Molar Mass of solvent

Vapour pressure of pure water at \(100\,{\,^o}C\)

(by assumption \(=760\,torr\) )

By substituting values in equation \((i)\) we get,

\(\frac{760-P_{x}}{P_{x}}=\frac{18 \times 18}{180 \times 178.2} \quad \ldots(i i)\)

On solving \((i i)\) we get

\(P_{s}=752.4\,torr\)