TS EAMCET · Chemistry · Solutions
Vapour pressure in \(\mathrm{mm} \mathrm{Hg}\) of 0.1 mole of urea in \(180 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\) is (The vapour pressure of water at \(25^{\circ} \mathrm{C}\) is \(24 \mathrm{~mm} \mathrm{Hg}\) )
- A \(2.376\)
- B \(20.76\)
- C \(23.76\)
- D \(24.76\)
Answer & Solution
Correct Answer
(C) \(23.76\)
Step-by-step Solution
Detailed explanation
From Raoult's law for very dilute solution. \[ \begin{aligned} \frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w}{m} \times \frac{M}{W} \\ \frac{24-p_s}{24} & =0.1 \times \frac{18}{180} \\ 24-p_s & =0.24 \\ \therefore \quad p_s= & 24-0.24=23.76 \mathrm{~mm} \mathrm{Hg} \end{aligned} \]
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