AP EAMCET · Chemistry · Solutions
When \(25 \mathrm{~g}\) of a non-volatile solute is dissolved in \(100 \mathrm{~g}\) of water, the vapour pressure is lowered by \(2.25 \times 10^{-1} \mathrm{~mm}\). If the vapour pressure of water at \(20^{\circ} \mathrm{C}\) is \(17.5 \mathrm{~mm}\), what is the molecular weight of the solute?
- A \(206\)
- B \(302\)
- C \(350\)
- D \(276\)
Answer & Solution
Correct Answer
(C) \(350\)
Step-by-step Solution
Detailed explanation
Given, weight of non-volatile solute, \[ w=25 \mathrm{~g} \] Weight of solvent, \(W=100 \mathrm{~g}\) Lowering of vapour pressure, \[ p^{\circ}-p_s=0.225 \mathrm{~mm} \] Vapour pressure of pure solvent, \[ p^{\circ}=17.5 \mathrm{~mm} \] Molecular weight of solvent…
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