   Chapter 12, Problem 12.116QP General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

A sample of an ionic solid is dissolved in 1.00 kg of water. The freezing point of the water is −0.01°C. If three times the mass of ionic solid is dissolved in 1.00 kg of water and the resulting freezing point of the solution is −0.09°C, which of the following would be possible formulas for the solid: MX, MX2, or MX3, where M represents a metal cation and X an anion with a l-charge? Explain your reasoning.

Interpretation Introduction

Interpretation:

Given the freezing point of the solution of ionic solid in water, the possible molecular formula of the ionic compound that cause the given depression in freezing point when mass of the compound is tripled has to be determined from one of the following - MX, MX2 or MX3 where M is metal cation and X is an anion with charge 1.

Concept Introduction:

Freezing point of a liquid substance is defined as the temperature at which the liquid becomes and solid and both the phases remain in equilibrium.

Depression in freezing point of a solvent due to the addition of non-volatile solute can be determined by the formula,

ΔTf = i×Kf×m .

Where,

ΔTf = depression in freezing pointKf  = depression in freezing point constantm    = molality of the solution; i = Van't Hoff factor

Explanation

van't Hoff factor 'i' refers to the number of ions produced per unit formula.  Accordingly the compounds MX, MX2 and MX3 produces 2,3,4 ions respectively, when dissolved in water.  We have to identify the compound which could cause the freezing point depression change from 0.01°C to 0.03°C.

For MX, i=2 . Therefore molality of the solution if the solute is MX is,

ΔTf = i×Kf×mm = ΔTfi×Kf = 0.01°C2×1.86°C/m = 2.69×103m

Mass of MX is tripled.  Then molality also gets tripled as 3×2.69×103m  .  Depression in freezing point for this molality is,

ΔTf = i×Kf×m = 2×1.86°C/m×3×2.69 m = 0.03°C

Hence the compound could be MX.

For MX2, i=3 . Therefore molality of the solution if the solute is MX2 is,

ΔTf = i×Kf×mm = ΔTfi×Kf = 0

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