Question

An aqueous soulution contains 30% C3H7OH and 70% water by mass. what are the mole fractions of each substance in the solution?

Answer 1

Answer:

Mole fraction of water = 0.88

Mole fraction of $\chi_{C_{3}H_{7}OH}$ = 0.12

Explanation:

30% $C_{3}H_{7}OH$ means 30 g of $C_{3}H_{7}OH$ is present in 100 g of solution.

70% $H_{2}O$(water) means 70 g of water is present in 100 g of solution

$n=\frac{given\ mass}{molar\ mass}$

Calculation of MOLES

Molar mass of  $H_{2}O$ = 18 g/mol

$n_{water}=\frac{70}{18}$

$n_{water} = 3.89\ moles$

moles of water = 3.89 moles

Molar mass of  $C_{3}H_{7}OH$ = 3(12) + 7(1) + 1(16) +1(1)

                                                                  = 36 +7+16+1

molar mass of $C_{3}H_{7}OH$ = 60 g/mol

$n_{C_{3}H_{7}OH}=\frac{30}{60}$

$n_{C_{3}H_{7}OH} = 0.5\ moles$

moles of $C_{3}H_{7}OH$ = 0.5 moles

total moles of solution = 0.5 + 3.89 = 4.39 moles

Calculation of MOLE FRACTION

Formula for mole fraction is :

$\chi=\frac{moles\ of\ solute}{moles\ of\ solution}$

$\chi_{water}=\frac{moles\ of\ water}{Total\ moles}$

$\chi_{water}=\frac{3.89}{4.39}$

$\chi_{water}= 0.88$

Similarly,

$\chi_{C_{3}H_{7}OH} = \frac{moles\ of\ C_{3}H_{7}OH}{Total\ moles}$

$\chi_{C_{3}H_{7}OH} = \frac{0.5}{4.39}$

$\chi_{C_{3}H_{7}OH} = \frac{0.5}{4.39}$

$\chi_{C_{3}H_{7}OH} = 0.12$

Mole fraction of water = 0.88

Mole fraction of $\chi_{C_{3}H_{7}OH}$ = 0.12

Note :

$\chi_{C_{3}H_{7}OH} + \chi_{H_{2}O} = 1$