(a) Explanation Given Density of acetonitrile ( CH 3 CN ) is 0.786 g/mL . Density of methanol ( CH 3 OH ) is 0.791 g/mL . Volume of acetonitrile ( CH 3 CN ) dissolved is 98.7 mL . Volume of methanol ( CH 3 OH ) dissolved is 22.5 mL . Mass of methanol in the solution is calculated as, m CH 3 OH = d CH 3 OH × V CH 3 OH Where, m CH 3 OH is the mass of methanol. d CH 3 OH is the density of methanol. V CH 3 OH is the volume of methanol. Substitute the values of d CH 3 OH and V CH 3 OH in the above formula. m CH 3 OH = 0.791 g/mL × 22.5 mL = 17.798 g Molar mass of methanol ( CH 3 OH ) is, 12.01 g/mol + 4 × 1.01 g/mol + 16.0 g/mol = 32.05 g/mol Moles of methanol are calculated as, n CH 3 OH = m CH 3 OH M CH 3 OH Where, n CH 3 OH is the number of moles of CH 3 OH . M CH 3 OH is the molar mass of CH 3 OH . Substitute the values of M CH 3 OH and m CH 3 OH in the above formula. n CH 3 OH = 17.798 g 32.05 g/mol = 0.555 mol Mass of acetonitrile in the solution is calculated as, m CH 3 CN = d CH 3 CN × V CH 3 CN Where, m CH 3 CN is the mass of acetonitrile (b) Interpretation Introduction To determine: The molality of the solution. (c) Interpretation Introduction To determine: The molarity of CH 3 OH in the solution.

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Author: Brown

Publisher: PEARSON CO

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Chapter 13, Problem 49E

(a)

Interpretation Introduction

To determine: The mole fraction of methanol in the solution.

(b)

Interpretation Introduction

To determine: The molality of the solution.

(c)

Interpretation Introduction

To determine: The molarity of CH3OH in the solution.

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Chapter 13, Problem 48E

Chapter 13, Problem 50E

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