50 mL of 0.2 molal urea solution (density = | vedclass

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Question

(A) Weight of $50 \ mL$ $0.2$ molal urea solution $= V 	imes d = 50 	imes 1.012 = 50.6 \ g$. Given $0.2$ molal implies $1000 \ g$ solvent has $0.2$

Vedclass · Accepted Answer

$682$

Vedclass · Answer

(A) Weight of $50 \ mL$ $0.2$ molal urea solution $= V 	imes d = 50 	imes 1.012 = 50.6 \ g$. Given $0.2$ molal implies $1000 \ g$ solvent has $0.2$ moles of urea. Weight of solution $= 1000 + (0.2 	imes 60) = 1012 \ g$. Weight of urea in $50.6 \ g$ solution $= \frac{12 	imes 50.6}{1012} = 0.6 \ g$. Total urea $= 0.6 + 0.06 = 0.66 \ g$. Total volume $= 50 \ mL + 250 \ mL = 300 \ mL = 0.3 \ L$. Osmotic pressure $\pi = C 	imes R 	imes T = \frac{n}{V} 	imes R 	imes T = \frac{0.66 / 60}{0.3} 	imes 62.36 	imes 300 \approx 686 \ Torr$. Using $R = 62.36 \ L \ Torr \ K^{-1} \ mol^{-1}$,the value is approximately $686 \ Torr$. Given the options,$682$ is the closest match.

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