   Chapter 13, Problem 12PS

Chapter
Section
Textbook Problem

Silver ion has an average concentration of 28 ppb (parts per billion) in U.S. water supplies. (a) What is the molality of the silver ion? (b) If you wanted 1.0 × 102 g of silver and could recover it chemically from water supplies, what volume of water in liters would you have to treat? (Assume the density of water is 1.0g/cm3.)

(a)

Interpretation Introduction

Interpretation: The molality of silver ions has to be identified.

Concept introduction:

Molality: The amount of solute (mol) per kilogram of solvent.

Concentration (C, mol/kg) =  molality of solute = Amount of solute (mol)Mass of solvent (Kg)

Explanation

Given data:

Concentrationofsilverions=28ppb

1 ppm = 1000 ppb = 281000ppm=0.028ppm

1ppm=1mg/kg,0.028 ppm = 0.028 mg/kg = 0.028 ×10-3g/kg= 0.028 ×10-3g/L(massofsilver ionsin1Lofwater)

No

(b)

Interpretation Introduction

Interpretation: The volume of water in litres has to be identified.

Concept introduction:

Molality: The amount of solute (mol) per kilogram of solvent.

Concentration (C, mol/kg) =  molality of solute = Amount of solute (mol)Mass of solvent (kg)

Density=massvolume

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