# If $1.02$ grams of urea when dissolved in $98.5$ grams of certain solvent decreases its freezing point by $0.211$ Kelvin. $1.60$ grams of an unknown compound when dissolved in $86.0$ grams of certain solvent depresses the freezing point by$0.34$Kelvin. Calculate the molar mass of the unknown compound. (Urea is ${\text{N}}{{\text{H}}_{\text{2}}}{\text{CON}}{{\text{H}}_{\text{2}}}$, where N = 14, H = 1, C = 12, O = 16)

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Hint:When a solute is dissolved in a solvent to form a solution, then the freezing point of the solution gets lowered in comparison to the freezing point of the solvent and this phenomenon is called the Depression in the freezing point of the solvent which depends on the number of moles of the solute present.

The depression in the freezing of a solution can be mathematically stated as,
${{\Delta }}{{\text{T}}_{\text{f}}}{\text{ = }}{{\text{k}}_{\text{f}}}{\text{ m}}$,
where ${{\text{k}}_{\text{f}}}$is the ebullioscopic constant of the solvent and m is the molality of the solution.
Given that,
The mass of urea = $1.02$ grams, molecular mass of urea = $\left[ {\left( {14 \times 2} \right) + 12 + 16 + \left( {1 \times 4} \right)} \right] = 60$
Therefore, the number of the moles of urea involved in the reaction = $\dfrac{{1.02}}{{60}}$
The molality of the solution is the number of moles dissolved per Kg of the solution hence it can be written as, $\dfrac{{1.02}}{{60}} \times \dfrac{{1000}}{{98.5}}$
= $0.17259$(m).
Putting the value of this molality in the above equation we get,
${\text{0}}{\text{.211 = }}{{\text{k}}_{\text{f}}}{\text{ 0}}{\text{.17259}}$………. Equation (1)
Now let the molar mass of the unknown compound be M. now, putting this value in the above equation we get,
${{\Delta }}{{\text{T}}_{\text{f}}}{\text{ = }}{{\text{k}}_{\text{f}}}{\text{ }}\dfrac{{{\text{1}}{\text{.60}}}}{{\text{M}}}{{ \times }}\dfrac{{{\text{1000}}}}{{{\text{80}}{\text{.6}}}}$$\Rightarrow 0.34{\text{ = }}{{\text{k}}_{\text{f}}}{\text{ }}\dfrac{{{\text{1}}{\text{.60}}}}{{\text{M}}}{{ \times }}\dfrac{{{\text{1000}}}}{{{\text{86}}{\text{.0}}}}$$ \Rightarrow 0.34{\text{ }} \times {\text{M = }}{{\text{k}}_{\text{f}}}{\text{ 19}}{\text{.85}}$………… Equation (2)
Dividing, equation (2) by equation (1) we get,
$\dfrac{{{\text{0}}{{.34 \times M}}}}{{{\text{0}}{\text{.211}}}}{\text{ = }}\dfrac{{{{\text{k}}_{\text{f}}}{\text{ 19}}{\text{.85}}}}{{{{\text{k}}_{\text{f}}}{\text{ 0}}{\text{.17259}}}}$.

Therefore, the molecular weight of the unknown compound is, ${\text{M = }}\dfrac{{{\text{19}}{\text{.85}} \times {\text{0}}{\text{.211}}}}{{{\text{0}}{\text{.17259}} \times {\text{0}}{\text{.34}}}}$= 70 grams.

Note:
The depression in the freezing point of the solvent is a colligative property which are properties that do not depend on the property or the nature of the solute particles but depend only on the number of moles of the particles present in the solution.