x gm of urea in 90 gm of water, shows same lowering of vapour pressure as 18gm of glucose in 90gm of water. Thus x is _____

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x gm of urea in 90 gm of water, shows same lowering of vapour pressure as 18gm of glucose in 90gm of water. Thus x is ______

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answer is 6.

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Detailed Solution

${(\frac{Δ P}{P^{0}})}_{U r e a} = x_{U r e a} = \frac{x / 60}{\frac{x}{60} + \frac{90}{18}}$

${(\frac{Δ P}{P^{0}})}_{g l u \cos e} = x_{g l u \cos e} = \frac{18 / 180}{\frac{18}{180} + \frac{90}{18}}$

Given ${(\frac{Δ P}{P^{0}})}_{U r e a} = {(\frac{Δ P}{P^{0}})}_{G l u c o s e}$

$∴ \frac{x / 60}{(\frac{x}{60} + 5)} = \frac{0.1}{0.1 + 5} = \frac{1}{51}$

$51 \times \frac{x}{60} = \frac{x}{60} + 5$

$\frac{50 x}{60} = 5$

$x = 6 g m$

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