Chemistry and Chemical Reactivity (9th Edition)

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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