Answer
410.1k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)