(a) Interpretation Introduction To determine: The mass ( g ) of ethanol needed to make aqueous solution of 6 .86 M ethanol. Explanation Given Volume of solution is 1 L . Molarity of ethanol solution is 6 .86 M . Molecular formula of ethanol is CH 3 CH 2 OH . Atomic mass of carbon is 12 .01 g/mol . Atomic mass of oxygen is 16 .00 g/mol . Atomic mass of hydrogen is 1 .01 g/mol . The number of moles of ethanol present in 6 .86 M solution is calculated by the formula, Moles of ethanol = Molarity × Volume of solution Substitute the value of molarity and volume of solution in the above equation. Moles of ethanol = 6 .86 M × 1 L = 6 .86 mol The molar mass of ethanol is calculated by the formula, Molar mass of ethanol = ( N C × M C ) + ( N O × M O ) + ( N H × M H ) Where, N C is the number of carbon atoms present in ethanol molecule. N O is the number of oxygen atoms present in ethanol molecule. N H is the number of hydrogen atoms present in ethanol molecule. M C is the mass of carbon atom. M O is the mass of oxygen atom (b) Interpretation Introduction To determine: The volume of ethanol needed to make 1 .00 L of vodka solution.

Chemistry: The Central Science (14th Edition)

14th Edition

ISBN: 9780134414232

Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

Publisher: PEARSON

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Chapter 4, Problem 67E

(a)

Interpretation Introduction

To determine: The mass (g) of ethanol needed to make aqueous solution of 6.86 M ethanol.

(b)

Interpretation Introduction

To determine: The volume of ethanol needed to make 1.00 L of vodka solution.

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Chapter 4, Problem 66E

Chapter 4, Problem 68E

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Chemistry: The Central Science (14th Edition)

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