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)
Answer
175.5k+ views
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.
Complete step by step answer:
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.
Complete step by step answer:
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
