When 5 g of a non-electrolyte solute is diss | vedclass

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(A) According to Raoult's Law for a non-volatile solute,the relative lowering of vapor pressure is given by: $\frac{P^o - P_s}{P^o} = \frac{w_2 	imes

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(A) According to Raoult's Law for a non-volatile solute,the relative lowering of vapor pressure is given by: $\frac{P^o - P_s}{P^o} = \frac{w_2 	imes M_1}{M_2 	imes w_1}$ Given: $P^o = 3000 \ N \ m^{-2}$ $P_s = 2985 \ N \ m^{-2}$ $w_2 = 5 \ g$ (mass of solute) $w_1 = 100 \ g$ (mass of solvent,water) $M_1 = 18 \ g/mol$ (molar mass of water) Substituting the values: $\frac{3000 - 2985}{3000} = \frac{5 	imes 18}{M_2 	imes 100}$ $\frac{15}{3000} = \frac{90}{M_2 	imes 100}$ $0.005 = \frac{0.9}{M_2}$ $M_2 = \frac{0.9}{0.005} = 180 \ g/mol$ Therefore,the molar mass of the solute is $180 \ g/mol$.

When $5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the vapor pressure decreases from $3000 \ N \ m^{-2}$ to $2985 \ N \ m^{-2}$. The molar mass of the solute is ............... $g/mol$.

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