HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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Solution - 1.0 x10-3Kg of urea when dissolved in 0.0985 Kg of a solvent, decreases freezing point of the solvent by 0.211 k. - HSC Science (Electronics) 12th Board Exam - Chemistry

Question

1.0 x10-3Kg of urea when dissolved in 0.0985 Kg of a solvent, decreases freezing point of the solvent by 0.211 k. 1.6x10 Kg of another non-electrolyte solute when dissolved in 0.086 Kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)

Solution

Mass of solute (A), i.e. urea = 1.0 × 10-3 kg

Molar mass of solute (A) = 60

Mass of solvent in which solute A is dissolved = 0.0985 kg

Mass of solute (B) = 1.6 × 10-3 kg

Molar mass of solute (B) = ?

Mass of solvent in which solute B is dissolved = 0.086 kg

`(DeltaT_(f_A))/(DeltaT_(f_B))=m_A/m_B`

`0.211/0.34=("Mass of solute (A)"/"Molecular mass of solute(A)×Kg of solvent")/("Mass of solute(B)"/"Molecular mass of solute(B)×Kg of solvent")`

`0.211/0.34=((1xx10^-3)/(60xx0.0985))/((1.6xx10^-3)/("Molecular mass of solute(B)" xx 0.086)`

`0.211/0.34=(1xx10^-3xx "Molecular mass of solute(B)" xx 0.086)/(60 xx 0.0985 xx 1.6 xx 10^-3)`

`"Molecular mass of solute(B)" = (0.211 xx 60 xx 0.0985 xx 1.6 xx 10^-3)/(0.34 xx 1 xx 10^-3 xx 0.086)`

Molar mass of another solute = 68.24

 

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Reference Material

Solution for question: 1.0 x10-3Kg of urea when dissolved in 0.0985 Kg of a solvent, decreases freezing point of the solvent by 0.211 k. concept: null - Depression of Freezing Point. For the courses HSC Science (Electronics), HSC Science (Computer Science), HSC Science (General)
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