A mixture of $N_{2}$ and water vapours is admitted into a flask which contains a solid drying agent. Immediately after admission, the pressure of the flask is 760 $t o r r$ . After standing some hours, the pressure reaches a steady value of 745 $t o r r$ .
If the experiment is conducted at $20^{o} C$ and the drying agent increases its mass by 0.15 $g$ , the volume becomes $X \times 10$ litre. The value of X is _______.

[Neglect volume occupied by drying agent.]

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Solution

$P_{H_{2} O} = 760 - 745 = 15 t o r r$

$w_{H_{2} O} = 0.15 g$

Moles of $H_{2} O = \frac{0.15}{18}$

For $H_{2} O (v)$ ,

$P V = n R T$

$\frac{15}{760} \times V = \frac{0.15}{18} \times 0.0821 \times 293$

$V = 10.16 l i t r e$

$X \times 10 = 10$

$∴ X = 1$

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PH2O=760−745=15 torrwH2O=0.15gMoles of H2O=0.1518For H2O(v), PV=nRT15760×V=0.1518×0.0821×293V=10.16litreX×10=10∴X=1

$P_{H_{2} O} = 760 - 745 = 15 t o r r$

$w_{H_{2} O} = 0.15 g$

Moles of $H_{2} O = \frac{0.15}{18}$

For $H_{2} O (v)$ ,

$P V = n R T$

$\frac{15}{760} \times V = \frac{0.15}{18} \times 0.0821 \times 293$

$V = 10.16 l i t r e$

$X \times 10 = 10$

$∴ X = 1$