Suppose `5 g` of acetic acid are dissolved in one litre of ethanol. Assume no reaction in between them. Calculate molality of resulting solution if density of ethanol is `0.789//mL`.

Options

A. ` 0.0856 `
B. ` 0.0956 `
C. ` 0.1056 `
D. ` 0.1156 `

Answer

C

Explanation

Wt. of `CH_(3)COOH`dissolved = `5g`
Eq. of `CH_(3)COOH` dissolved =`(5)/(60)`
Volume of ethanol =`1` litre `1000ml.`
`:.`Weight of ethanol =`1000xx0.789 =789 g`
`:.`Weight of ethanol =`1000xx0.789=789g`
`:.`Molartiy of solution
`=("Moles of solute")/("wt. of solvent in kg")=((5)/(60xx789))/(1000)=0.1056`