   Chapter 13, Problem 82GQ

Chapter
Section
Textbook Problem

A compound is known to be a potassium halide, KX. If 4.00 g of the salt is dissolved in exactly 100 g of water, the solution freezes at −1.28 °C. Identity the halide ion in this formula.

Interpretation Introduction

Interpretation: The halide ion in KX 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 potassium halide is determined.

Given,

ΔT=1.280C

Mass of potassium halide is 4g

Mass of water is 100g=0.1kg

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

Molar mass of potassium is 39.10g/mol

Change in freezing point is calculated by using the equation,

ΔTbp=Kbpmsolute

Hence,

The concentration is calculated by,

Concentration,msolute=ΔTbpKbp=(1.280C)(1.860C/m)=0.6882m

The amount of potassium halide is,

(0.6882m1kg)(0

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