Calculate the relative lowering of vapour pre | vedclass

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(B) The relative lowering of vapour pressure is given by Raoult's Law as: $\frac{P^o - P_s}{P^o} = \chi_{solute} = \frac{n_2}{n_1 + n_2}$.Given: Mass

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$0.0017$

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(B) The relative lowering of vapour pressure is given by Raoult's Law as: $\frac{P^o - P_s}{P^o} = \chi_{solute} = \frac{n_2}{n_1 + n_2}$.Given: Mass of solute $(w_2)$ $= 0.56 \ g$,Molar mass of solute $(M_2)$ $= 60 \ g \ mol^{-1}$,Mass of solvent $(w_1)$ $= 100 \ g$,Molar mass of water $(M_1)$ $= 18 \ g \ mol^{-1}$.Moles of solute $(n_2)$ $= \frac{0.56}{60} \approx 0.00933 \ mol$.Moles of solvent $(n_1)$ $= \frac{100}{18} \approx 5.556 \ mol$.Since $n_2$ is very small compared to $n_1$,we can approximate $\frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$.Relative lowering of vapour pressure $= \frac{0.00933}{5.556} \approx 0.00168 \approx 0.0017$.

Calculate the relative lowering of vapour pressure of a solution containing $0.56 \ g$ of a non-volatile solute in $100 \ g$ of water [molar mass of solute $= 60 \ g \ mol^{-1}$].

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