A 3.5 molal aqueous solution of methyl alcohol `(CH_(3)OH)` is supplied. What is the mole fraction of methyl alcohol in the solution ?

Options

A. ` 0.100 `
B. ` 0.059 `
C. ` 0.086 `
D. ` 0.050 `

Answer

B

Explanation

`("Moles of (solute)" CH_(3)OH)/(1 kg "of" H_(2)O) =(3.5 mol)/(1 kg)`
Thus ,`n(CH_(3)OH)=3.5`
`N(H_(2)O)=(1000)/(18)=55.56 `
`:.` Total moles `= 3.5+55.56 =59.06 mol`
`:. chi_(CH_(3)OH) =("Moles of" CH_(3)OH)/("Total Moles") =(3.5)/(59.05)=0.059`