A 3.5 molal aqueous solution of methyl alcohol (CH3OH) is supplied. What is the mole fraction of methyl alcohol in the solution?

Question

A 3.5 molal aqueous solution of methyl alcohol (CH3OH) is supplied. What is the mole fraction of methyl alcohol in the solution? (A) 0.100 (B) 0.059 (C) 0.086 (D) 0.050

Tardigrade Admin · Accepted Answer

( textMoles of(solute) CH3 OH/ text1kg of H2O) = (3.5 mol/1kg) Thus, n(CH3OH)=3.5 N(H2O)=(1000/18)=55.56 ∴ Total moles =3.5+55.56=59.06 mol ∴ chiCH3OH =( textMoles of CH3OH/ textTotal moles) =(3.5/59.05)=0.059

Q. A 3.5 molal aqueous solution of methyl alcohol $(C H_{3} O H)$ is supplied. What is the mole fraction of methyl alcohol in the solution?

$0.100$

$0.059$

$0.086$

$0.050$

$\frac{Moles of(solute) C H _{3} OH}{1kg of H _{2} O} = \frac{3.5 m o l}{1 k g}$
Thus, $n (C H_{3} O H) = 3.5$
$N (H_{2} O) = \frac{1000}{18} = 55.56$
$∴$ Total moles $= 3.5 + 55.56 = 59.06 m o l$
$∴ χ_{C H_{3} O H} = \frac{Moles of C H _{3} OH}{Total moles}$
$= \frac{3.5}{59.05} = 0.059$

Q. A 3.5 molal aqueous solution of methyl alcohol (CH3​OH) is supplied. What is the mole fraction of methyl alcohol in the solution?

0.100

0.059

0.086

0.050