The vapour pressure of water at room temperat | vedclass

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(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^0 - P_s}{P^0} = \chi_B = 0.05$. Here,$\chi_B$ is the mole fraction of t

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$3$

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(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^0 - P_s}{P^0} = \chi_B = 0.05$. Here,$\chi_B$ is the mole fraction of the solute. Since $\chi_B = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A} = 0.05$ (for dilute solutions),where $n_A$ is the number of moles of water. $n_A = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$. $n_B = 0.05 	imes 55.55 = 2.777 \ mol$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. $m = \frac{n_B 	imes 1000}{W_A 	ext{ (in g)}} = \frac{2.777 	imes 1000}{1000} = 2.777 \ m \approx 3 \ m$.

The vapour pressure of water at room temperature is lowered by $5\%$ by dissolving a solute in it. Then the approximate molality of the solution is

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