Question

What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass

• A. $0.69$
• B. $0.82$
• C. $0.85$
• D. $0.60$

Answer 1

Answer: (A)

$0.69$

Answer 2

$\mathrm { x } _ { \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } } = \frac { \mathrm { n } _ { \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } } } { \mathrm { n } _ { \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } } + \mathrm { n } _ { \mathrm { H } _ { 5 } \mathrm { O } } }$

$\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }$ dk nzO;eku = 85 g

Mass of $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } = 46 \mathrm {~g} / \mathrm { mol }$

$\mathrm { n } _ { \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } } = \frac { 85 } { 46 } = 1.85 \mathrm {~mol}$

Mass water = 100 -85 = 15g

$\mathrm { n } _ { \mathrm { H } _ { 2 } \mathrm { O } } = \frac { 15 } { 18 } = 0.833 \mathrm {~mol}$

$\mathrm { x } _ { \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } } = \frac { 1.85 } { 1.85 + 0.833 } = \frac { 1.85 } { 2.683 } = 0.69$