HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
Share

# A solution containing 0.73 g of camphor (molar mass 152 g . mol-1) in 36.8 g of acetone (boiling point 56.3°C) boils at 56.55°C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46oC. Calculate the molar mass of the unknown compound. - HSC Science (Computer Science) 12th Board Exam - Chemistry

ConceptColligative Properties and Determination of Molar Mass - Introduction

#### Question

A solution containing 0.73 g of camphor (molar mass 152 g . mol-1) in 36.8 g of acetone (boiling point 56.3°C) boils at 56.55°C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46oC. Calculate the molar mass of the unknown compound.

#### Solution

Calculation for Kb of camphor:

Here, mass of solvent, WA = 36.8g

mass of solute, WB = 0.73g

molecular mass of solute, MB = 152

elevation in boiling point,

ΔTb = 56.55 - 56.30 = 0.25°C

Now,

ΔTb = Kb × (WB × 1000)/(WA × MB)

Kb = (ΔTb × WA × MB)/(WB × 1000)

Kb = (0.25 × 36.8 × 152)/(0.73 × 1000)

Kb =1.9156 °C kgmol-1

Calculation of molecular mass of unknown solute

Here, mass of solvent, WA = 36.8g

mass of solute, WB = 0.564g

elevation in boiling point,

ΔTb = 56.46 - 56.30 = 0.16°C

Now,

Mb = (Kb × WB × 1000)/(WA × ΔTb)

Mb = (1.9156 × 0.564 × 1000)/(36.8 × 0.16)

Mb =183.4 g

Is there an error in this question or solution?

#### APPEARS IN

2013-2014 (October) (with solutions)
Question 2.4 | 3.00 marks

#### Video TutorialsVIEW ALL 

Solution A solution containing 0.73 g of camphor (molar mass 152 g . mol-1) in 36.8 g of acetone (boiling point 56.3°C) boils at 56.55°C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46oC. Calculate the molar mass of the unknown compound. Concept: Colligative Properties and Determination of Molar Mass - Introduction.
S