Question

Calculate the mole fraction of ethylene glycol (C2H6O2,) and water in a solution containing
20% of C2H603, by mass.
​

Answer 1

Answer :

• Mole fraction of ethylene glycol in the solution = 0.068
• Mole fraction of water in the solution = 0.932

Step-by-step explanation :

Given that,

• 20 g of ($\bold{C_2H_6O_2}$) is present in 100 g of the solution.
• Mass of solute ($\bold{C_2H_6O_2}$) = 20 g

$\hrulefill$

Now,

• Mass of solvent ($\bold{H_20}$) = 100 - 20 g = 80 g
• Molar mass of $\bold{C_2H_6O_2=62\:g\:mol^{-1}}$
• Molar mass of $\bold{H_2O=18\:g\:mol^{-1}}$

∴ No. of moles of $\bold{C_2H_6O_2=\frac{Mass\:in\:g}{Molar\:mass}}$

$\implies\bold{\dfrac{20}{62}=0.322}$

No. of moles of $\bold{H_2O=\dfrac{80}{18}=4.444}$

∴ Mole fraction of $\bold{C_2H_6O_2}$ in the solution,

$\implies\bold{\dfrac{n_{C_2H_6O_2}}{n_{C_2H_6O_2}+n_{H_2O}}}$

$\implies\bold{\dfrac{0.322}{0.322+4.444}}$

$\implies\bold{0.068}$

So, mole fraction of $\bold{H_2O}$ in the solution,

$\implies\bold{1-0.068}$

$\implies\bold{0.932}$

Answer 2

20% of C2H603, by mass

Means 20g of C2H603 in 80g of water

Mass of C2H602 =20g

Molar mass

=2(C)+6(H)+3(O)

=2(12)+6(1)+2(16)

=62

Moles (C2H602)=20/62=0.32

Mass of water=80g

Molar mass of water=18g

Moles of water=80/18=4.44

Total moles

= moles C2H602+moles of water

=0.32+4.44

=4.76

Mole fraction C2H602

= Moles C2H602/total moles

=0.32/4.76

=0.067

Mole fraction of water

=1-0.067

=0.933

I hope this helps ( ╹▽╹ )