Explanation Given, A 9 .360-g sample of the window cleaner was diluted with 35 .778 g water and 4 .188 g of this diluted solution was titrated using 13 .58 mL of 0 .1093-M HCl . Total mass of the dilute solution is given below, The total mass of the dilute solution = 9 .360 g WC + 35 .778 g water The total mass of the dilute solution = 45 .138 g dilute solution . Molar Mass of NH 3 = 17 .0304 g/mol Mass of ammonia = 13 .58 mL × 1L 1000 ml × 0 .1093 mol HCl 1 L × 1 mol NH 3 1 mol HCl × 17

5th Edition

ISBN: 9781285199047

Author: John W. Moore, Conrad L. Stanitski

Publisher: Cengage Learning

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

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Chapter 3, Problem 154QRT

Interpretation Introduction

Interpretation:

A 9.360-g sample of the window cleaner was diluted with 35.778 g water and 4.188 g of this diluted solution was titrated using 13.58 mL of 0.1093-M HCl. The mass percent of ammonia has to be calculated.

Concept introduction:

The mass percent can be calculated by using following formula

Mass percent = Actual mass of product Theoretical mass of product×100

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Chapter 3, Problem 153QRT

Chapter 3, Problem 155QRT

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Solution Summary: The author explains that the mass percent of ammonia has to be calculated by using formula Masspercent . The total mass of the dilute solution is given below.

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