   Chapter 13, Problem 45PS

Chapter
Section
Textbook Problem

An aqueous solution contains 0.180 g of an unknown, nonionic solute in 50.0 g of water. The solution freezes at −0.040 °C. What is the molar mass of the solute?

Interpretation Introduction

Interpretation: The molar mass of unknown nonionic solute has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Change in freezing point is calculated by using the equation,

ΔTfp=Kfpmsolute

where,

Kfp is the molal freezing point depression constant.

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Explanation

The molar mass of unknown nonionic solute is calculated.

Given,

ΔT=0.0400C

Mass of unknown nonionic solute is 0.180g

Mass of water is 50g=0.050kg

Molal freezing point depression constant of water is 1.860C/m

Change in freezing point is calculated by using the equation,

ΔTfp=Kfpmsolute

Hence,

The concentration is calculated by,

Concentration,msolute=ΔTfpKfp=(0.0400C)(1.860C/m)=0.0215m

The amount of unknown nonionic solute is,

(0

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