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Question

' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' $g{mol}^{- 1}$ when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg . If aqueous solution of the electrolyte boils at 375 K and $K_{b}$ for water is $0.52\text{ }K\text{ }kg{\text{ }mol}^{- 1}$, then the mole fraction of the electrolyte solute ( $x_{2}$ ) in the solution can be expressed as

(Given : density of water $= 1\text{ }g/mL$ and boiling point of water $= 373\text{ }K$ )

Options

A.$\frac{1.3}{8} \times \frac{W}{M}$
B.$\frac{16}{2.6} \times \frac{W}{M}$
C.$\frac{2.6}{16} \times \frac{M}{W}$
D.$\frac{1.3}{8} \times \frac{M}{W}$

Solution

$\ P^{\circ} = 640\text{ }mmHg$

$${P_{s} = 600\text{ }mmHg }{\Delta P = 40\text{ }mmHg }$$moles of solute $= \frac{W}{M}$

$$\frac{\Delta P}{P^{o}} = i.X_{\text{solute~}} $$Again :

$${\Delta T_{b} = i \times k_{b} \times m }{2 = i \times 0.52 \times \frac{W/M}{100} \times 1000 }{i = \frac{2}{5.2} \times \frac{M}{W} }{X_{\text{solute~}} = \frac{40}{640} \times \frac{1}{i} = \frac{1}{16} \times \frac{5.2}{2} \times \frac{W}{M} }{X_{\text{solute~}} = \frac{1.3}{8} \times \frac{W}{M}}$$

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