6 g urea is dissolved in 1 kg solvent. The | vedclass

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(A) $T_1 =$ Freezing point of pure solvent$T_2 =$ Freezing point of solution$\Delta T_f = T_1 - T_2 = i 	imes K_f 	imes m$For urea,van't Hoff factor

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$0.2$

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(A) $T_1 =$ Freezing point of pure solvent$T_2 =$ Freezing point of solution$\Delta T_f = T_1 - T_2 = i 	imes K_f 	imes m$For urea,van't Hoff factor $i = 1$.Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{6 \ g / 60 \ g \ mol^{-1}}{1 \ kg} = 0.1 \ mol \ kg^{-1}$.$\Delta T_f = 1 	imes 2 \ kg \ K \ mol^{-1} 	imes 0.1 \ mol \ kg^{-1} = 0.2 \ K$.Since the difference in temperature is the same in $^\circ C$ and $K$,$T_1 - T_2 = 0.2 \ ^\circ C$.

$6 \ g$ urea is dissolved in $1 \ kg$ solvent. The value of $(T_1 - T_2)$ will be .......... $^\circ C$. $(K_f = 2 \ kg \ K \ mol^{-1})$

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