JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(2.5 \mathrm{~g}\) of a non-volatile, non-electrolyte is dissolved in \(100 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\). The solution showed a boiling point elevation by \(2^{\circ} \mathrm{C}\). Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is _______ \(\mathrm{mm}\) of \(\mathrm{Hg}\) (nearest integer) [Given : Molal boiling point elevation constant of water \(\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{K.} \mathrm{kg} \mathrm{mol}^{-1}\), \(1 \mathrm{~atm}\) pressure \(=760 \mathrm{~mm}\) of \(\mathrm{Hg}\), molar mass of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\)
- A \(702\)
- B \(704\)
- C \(705\)
- D \(707\)
Answer & Solution
Correct Answer
(D) \(707\)
Step-by-step Solution
Detailed explanation
\( 2=0.52 \times \mathrm{m}\) \(\mathrm{m}=\frac{2}{0.52}\) According to question, solution is much diluted \(\text { so } \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{n}_{\text {solate }}}{\mathrm{n}_{\text {soltron: }}}\)…
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