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(A) The elevation in boiling point is given by $\Delta T_b = K_b 	imes m$. Given $\Delta T_b = 2 \ ^\circ C$,$K_b = 0.76 \ K \ kg \ mol^{-1}$,mass of

Vedclass · Accepted Answer

$724$

Vedclass · Answer

(A) The elevation in boiling point is given by $\Delta T_b = K_b 	imes m$. Given $\Delta T_b = 2 \ ^\circ C$,$K_b = 0.76 \ K \ kg \ mol^{-1}$,mass of solute $w_2 = 2.5 \ g$,and mass of solvent $w_1 = 100 \ g$. $2 = 0.76 	imes \frac{2.5 	imes 1000}{M_2 	imes 100}$. $M_2 = \frac{0.76 	imes 2.5 	imes 10}{2} = 9.5 \ g \ mol^{-1}$. Using Raoult's law for relative lowering of vapour pressure: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1} = \frac{w_2 	imes M_1}{M_2 	imes w_1}$. Here $P^\circ = 760 \ mm \ Hg$ and $M_1 = 18 \ g \ mol^{-1}$. $\frac{760 - P_s}{760} = \frac{2.5 	imes 18}{9.5 	imes 100} = \frac{45}{950} \approx 0.04737$. $760 - P_s = 760 	imes 0.04737 = 36$. $P_s = 760 - 36 = 724 \ mm \ Hg$.

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