Answer:
Mole fraction of water = 0.88
Mole fraction of [tex]\chi_{C_{3}H_{7}OH}[/tex] = 0.12
Explanation:
30% [tex]C_{3}H_{7}OH[/tex] means 30 g of [tex]C_{3}H_{7}OH[/tex] is present in 100 g of solution.
70% [tex]H_{2}O[/tex](water) means 70 g of water is present in 100 g of solution
[tex]n=\frac{given\ mass}{molar\ mass}[/tex]
Calculation of MOLES
Molar mass of [tex]H_{2}O[/tex] = 18 g/mol
[tex]n_{water}=\frac{70}{18}[/tex]
[tex]n_{water} = 3.89\ moles[/tex]
moles of water = 3.89 moles
Molar mass of [tex]C_{3}H_{7}OH[/tex] = 3(12) + 7(1) + 1(16) +1(1)
= 36 +7+16+1
molar mass of [tex]C_{3}H_{7}OH[/tex] = 60 g/mol
[tex]n_{C_{3}H_{7}OH}=\frac{30}{60}[/tex]
[tex]n_{C_{3}H_{7}OH} = 0.5\ moles[/tex]
moles of [tex]C_{3}H_{7}OH[/tex] = 0.5 moles
total moles of solution = 0.5 + 3.89 = 4.39 moles
Calculation of MOLE FRACTION
Formula for mole fraction is :
[tex]\chi=\frac{moles\ of\ solute}{moles\ of\ solution}[/tex]
[tex]\chi_{water}=\frac{moles\ of\ water}{Total\ moles}[/tex]
[tex]\chi_{water}=\frac{3.89}{4.39}[/tex]
[tex]\chi_{water}= 0.88[/tex]
Similarly,
[tex]\chi_{C_{3}H_{7}OH} = \frac{moles\ of\ C_{3}H_{7}OH}{Total\ moles}[/tex]
[tex]\chi_{C_{3}H_{7}OH} = \frac{0.5}{4.39}[/tex]
[tex]\chi_{C_{3}H_{7}OH} = \frac{0.5}{4.39}[/tex]
[tex]\chi_{C_{3}H_{7}OH} = 0.12[/tex]
Mole fraction of water = 0.88
Mole fraction of [tex]\chi_{C_{3}H_{7}OH}[/tex] = 0.12
Note :
[tex]\chi_{C_{3}H_{7}OH} + \chi_{H_{2}O} = 1[/tex]
