   Chapter 12, Problem 12.23QP 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

If 291g of a compound is added to 1.02 kg of water to increase the boiling point by 5.77°C, what is the molar mass of the added compound? (Assume a van ’t Hoff factor of 1.) a 31.1 g/mol b 30.3 g/mol c 28.5 g/mol d 18.3 g/mol e 25.3 g/mol

Interpretation:

Given the elevation of boiling point, the molecular weight of a compound has to be determined and the correct answer has to be chosen from the following options –

1. (a) 31.1 g/mol (b) 30.3 g/mol (c) 28.5 g/mol (d) 18.3 g/mol (e) 25.3 g/mol
Interpretation Introduction

Concept Introduction:

Elevation in boiling point is a colligative property which refers to increase in boiling point of the solution due to the addition of non-volatile solute. It is expressed as,

ΔTb = Kb.cm

Where,

ΔTb = elevation in boiling pointKb = boiling point elevation constantcm = molal concentration

Molality or molal concentration is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

Molality = number of moles of solutemass of solvent in kg

Explanation

Given that 291 g of a compound is dissolved in 1.02 kg of water.  Boiling point elevation of the solution, ΔTb = 5.77°C .

From the data given in the table 12.3 in text book, Kb for water is 0.512°C/m.

Therefore, molality of the solution is calculated as,

ΔTb = Kb.cm

Rewriting the above equation,

cm  ΔTbKb 5.77°C0.512°C/m = 11.27 m

As we know,

Molality  = number of moles of solutemass of solvent in kg

Rewriting the above expression,

no.of moles of solute  = molality × mass of solvent = 11

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