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(A) Let the mass of urea be $w \ g$. The molar mass of urea $(NH_2CONH_2)$ is $60 \ g \ mol^{-1}$.Molality $(m) = \frac{w 	imes 1000}{60 	imes 10000

Vedclass · Accepted Answer

$59.64$

Vedclass · Answer

(A) Let the mass of urea be $w \ g$. The molar mass of urea $(NH_2CONH_2)$ is $60 \ g \ mol^{-1}$.Molality $(m) = \frac{w 	imes 1000}{60 	imes 10000} = \frac{w}{600} \ mol \ kg^{-1}$.Elevation in boiling point: $\Delta T_b = K_b 	imes m = 0.513 	imes \frac{w}{600}$.Depression in freezing point: $\Delta T_f = K_f 	imes m = 1.86 	imes \frac{w}{600}$.The boiling point of the solution is $T_b = 100 + \Delta T_b$ and the freezing point is $T_f = 0 - \Delta T_f$.The difference is $(100 + \Delta T_b) - (0 - \Delta T_f) = 100 + \Delta T_b + \Delta T_f = 100.2372$.Therefore,$\Delta T_b + \Delta T_f = 0.2372$.$\frac{w}{600} 	imes (0.513 + 1.86) = 0.2372$.$\frac{w}{600} 	imes 2.373 = 0.2372$.$w = \frac{0.2372 	imes 600}{2.373} \approx 60 \ g$.Comparing with options,the closest value is $59.64 \ g$.

Column-$I$ (Various solutions)	Column-$II$ (Their freezing point)
$a$. $0.1 \ M \ BaCl_2$ solution	$p$. $271 \ K$
$b$. $0.1 \ M \ NaCl$ solution	$q$. $270 \ K$
$c$. $0.1 \ M \ K_3[Fe(CN)_6]$ solution	$s$. $269 \ K$
$d$. $0.1 \ M \ Al_2(SO_4)_3$ solution	$r$. $268 \ K$

The difference between the boiling point and freezing point of an aqueous solution of urea containing $10.0 \ kg$ of water is $100.2372 \ ^oC$. How many grams of urea are dissolved in the solution? (Given: $K_b = 0.513 \ K \ kg \ mol^{-1}$,$K_f = 1.86 \ K \ kg \ mol^{-1}$)

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