For a dilute solution containing 2.5 g of a | vedclass

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(A) The elevation in boiling point is given by $\Delta T_b = \frac{1000 	imes K_b 	imes w_2}{M_2 	imes w_1}$.Given $\Delta T_b = 2 \ K$,$K_b = 0.76

Vedclass · Accepted Answer

$724$

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(A) The elevation in boiling point is given by $\Delta T_b = \frac{1000 	imes K_b 	imes w_2}{M_2 	imes w_1}$.Given $\Delta T_b = 2 \ K$,$K_b = 0.76 \ K \ kg \ mol^{-1}$,$w_2 = 2.5 \ g$,$w_1 = 100 \ g$.$2 = \frac{1000 	imes 0.76 	imes 2.5}{M_2 	imes 100}$ $\Rightarrow M_2 = \frac{1000 	imes 0.76 	imes 2.5}{200} = 9.5 \ g \ mol^{-1}$.Using Raoult's law for relative lowering of vapor pressure: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1} = \frac{w_2 / M_2}{w_1 / M_1}$.Here $P^0 = 760 \ mm \ Hg$,$M_1 = 18 \ g \ mol^{-1}$.$\frac{760 - P_s}{760} = \frac{2.5 / 9.5}{100 / 18} = \frac{2.5 	imes 18}{9.5 	imes 100} = \frac{45}{950} \approx 0.04737$.$760 - P_s = 760 	imes 0.04737 = 36$.$P_s = 760 - 36 = 724 \ mm \ Hg$.

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