by
Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.

Published by
Cengage Learning

ISBN 10:
1133949649

ISBN 13:
978-1-13394-964-0

Chapter 13 Solutions and Their Behavior - Study Questions - Page 505: 3

Answer

COMPOUND C12H22O11
STEP 1:
Data given :
molality of C12H22O11 = 0.15
weight percent = ?
mole fraction = ?
STEP 2:
From molality it means that we have 0.15 moles of C12H22O11 dissolved in 1000 g of water
so moles of C12H22O11
Moles of water = 1000 \div 18
=55.5 g/mol
To find weight percent we need mass of C12H22O11
moles = mass in g \ molar mass
molar mass of C12H22O11
=12\times 12 + 1\times 22 + 16\times 11
= 342
mass in g = moles \times molar mass
= o.15 \times 342
=51.3
STEP 3:
weight percent = mass of C12H22O11 \div (mass of C12H22O11 + mass of water ) \times 100
=51.3 \div (51.3+ 1000) \times 100
=4.879 \approx 4.9 %
STEP 4:
mole fraction = moles of C12H22O11 \div ( moles of C12H22011 + moles of water )
= 0.15 \div ( 0.15 + 55.5 )
= 0.00269
RESULTS : weight percent = 4.9 % , mole fraction = 0.0269 .

Work Step by Step

COMPOUND C12H22O11
STEP 1:
Data given :
molality of C12H22O11 = 0.15
weight percent = ?
mole fraction = ?
STEP 2:
From molality it means that we have 0.15 moles of C12H22O11 dissolved in 1000 g of water
so moles of C12H22O11
Moles of water = 1000 \div 18
=55.5 g/mol
To find weight percent we need mass of C12H22O11
moles = mass in g \ molar mass
molar mass of C12H22O11
=12\times 12 + 1\times 22 + 16\times 11
= 342
mass in g = moles \times molar mass
= o.15 \times 342
=51.3
STEP 3:
weight percent = mass of C12H22O11 \div (mass of C12H22O11 + mass of water ) \times 100
=51.3 \div (51.3+ 1000) \times 100
=4.879 \approx 4.9 %
STEP 4:
mole fraction = moles of C12H22O11 \div ( moles of C12H22011 + moles of water )
= 0.15 \div ( 0.15 + 55.5 )
= 0.00269
RESULTS : weight percent = 4.9 % , mole fraction = 0.0269 .

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