A solution of 1.25 g of 'P' in 50 g of wate | vedclass

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(D) Given: Mass of solute $w_{B} = 1.25 \ g$,Mass of solvent $w_{A} = 50 \ g$,$\Delta T_{f} = 0.3 \ K$,$M_{	ext{normal}} = 94 \ g \ mol^{-1}$,$K_{f}

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None of the above

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(D) Given: Mass of solute $w_{B} = 1.25 \ g$,Mass of solvent $w_{A} = 50 \ g$,$\Delta T_{f} = 0.3 \ K$,$M_{	ext{normal}} = 94 \ g \ mol^{-1}$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$.First,calculate the observed molar mass $(M_{	ext{obs}})$ using the formula: $\Delta T_{f} = K_{f} 	imes m = K_{f} 	imes \frac{w_{B} 	imes 1000}{M_{	ext{obs}} 	imes w_{A}}$.$M_{	ext{obs}} = \frac{1.86 	imes 1.25 	imes 1000}{0.3 	imes 50} = 155 \ g \ mol^{-1}$.Calculate the van't Hoff factor $(i)$: $i = \frac{M_{	ext{normal}}}{M_{	ext{obs}}} = \frac{94}{155} \approx 0.6064$.For association,$n P ightleftharpoons P_{n}$. Assuming dimerization $(n=2)$,the degree of association $\alpha$ is given by $i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha(1 - \frac{1}{n})$.$0.6064 = 1 - \alpha(1 - 0.5) \implies 0.6064 = 1 - 0.5\alpha$.$0.5\alpha = 1 - 0.6064 = 0.3936$.$\alpha = 0.7872$ or approximately $78.7\%$.Since $78.7\%$ is not among the options,the correct choice is $D$.

$A$ solution of $1.25 \ g$ of '$P$' in $50 \ g$ of water lowers the freezing point by $0.3^{\circ}C$. The molar mass of '$P$' is $94 \ g \ mol^{-1}$. Given $K_{f(\text{water})} = 1.86 \ K \ kg \ mol^{-1}$,the degree of association of '$P$' in water is:

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