x moles of CO(NH_2)_2 are present in 1200 g | vedclass

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(B) The depression in freezing point is given by the formula: $\Delta T_f = k_f 	imes m$,where $m$ is the molality of the solution.$\Delta T_f = T_f^

Vedclass · Accepted Answer

$2.59$

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(B) The depression in freezing point is given by the formula: $\Delta T_f = k_f 	imes m$,where $m$ is the molality of the solution.$\Delta T_f = T_f^\circ - T_f = 0 \ ^oC - (-4.02 \ ^oC) = 4.02 \ K$.Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{x}{1.2 \ kg}$.Substituting the values into the formula: $4.02 = 1.86 	imes \frac{x}{1.2}$.Solving for $x$: $x = \frac{4.02 	imes 1.2}{1.86} = \frac{4.824}{1.86} \approx 2.59 \ 	ext{moles}$.

$x$ moles of $CO(NH_2)_2$ are present in $1200 \ g$ of water. If the freezing point of the solution is $-4.02 \ ^oC$,calculate the value of $x$. Given $k_f \ (H_2O) = 1.86 \ K \ kg \ mol^{-1}$.

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