The mole fraction of urea in an aqueous urea | vedclass

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(D) Moles of water $(n_{H_2O})$ = $\frac{900}{18} = 50 	ext{ mol}$.Mole fraction of urea $(x_{urea})$ = $0.05$.Using the formula $x_{urea} = \frac{n_

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$2.98$

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(D) Moles of water $(n_{H_2O})$ = $\frac{900}{18} = 50 	ext{ mol}$.Mole fraction of urea $(x_{urea})$ = $0.05$.Using the formula $x_{urea} = \frac{n_{urea}}{n_{urea} + n_{H_2O}}$:$0.05 = \frac{n_{urea}}{n_{urea} + 50}$.$0.05n_{urea} + 2.5 = n_{urea} \implies 0.95n_{urea} = 2.5 \implies n_{urea} = \frac{2.5}{0.95} \approx 2.6316 	ext{ mol}$.Mass of urea = $2.6316 	ext{ mol} 	imes 60 	ext{ g mol}^{-1} = 157.896 	ext{ g}$.Total mass of solution = $157.896 	ext{ g} + 900 	ext{ g} = 1057.896 	ext{ g}$.Volume of solution $(V)$ = $\frac{	ext{Mass}}{	ext{Density}} = \frac{1057.896 	ext{ g}}{1.2 	ext{ g cm}^{-3}} = 881.58 	ext{ mL} = 0.88158 	ext{ L}$.Molarity $(M)$ = $\frac{n_{urea}}{V(L)} = \frac{2.6316}{0.88158} \approx 2.98 	ext{ M}$.

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