The freezing point of a solution containing 1 | vedclass

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(D) Given: Volume of solute $A = 10 \ mL$ $\Delta T_f = 0 - (-0.413) = 0.413 \ K$ Mass of solvent $(w_A) = 500 \ g$ $K_f 	ext{ (water)} = 1.86 \ K \

Vedclass · Accepted Answer

$0.993$

Vedclass · Answer

(D) Given: Volume of solute $A = 10 \ mL$ $\Delta T_f = 0 - (-0.413) = 0.413 \ K$ Mass of solvent $(w_A) = 500 \ g$ $K_f 	ext{ (water)} = 1.86 \ K \ kg \ mol^{-1}$ Molecular weight of $A \ (M_B) = 60 \ g \ mol^{-1}$ Using the formula for freezing point depression: $\Delta T_f = K_f 	imes \frac{w_B}{M_B} 	imes \frac{1000}{w_A}$ $0.413 = 1.86 	imes \frac{w_B}{60} 	imes \frac{1000}{500}$ $	herefore w_B = \frac{0.413 	imes 60 	imes 500}{1.86 	imes 1000} = 6.66 \ g$ Total mass of solution $= w_A + w_B = 500 + 6.66 = 506.66 \ g$ Total volume of solution $= V_{	ext{solvent}} + V_{	ext{solute}} = 500 \ mL + 10 \ mL = 510 \ mL$ (Assuming density of water $\approx 1 \ g \ mL^{-1}$) Density of solution $(d) = \frac{	ext{Total mass}}{	ext{Total volume}} = \frac{506.66 \ g}{510 \ mL} \approx 0.993 \ g \ mL^{-1}$ Thus, the correct option is $(D)$.

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