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Suppose 5 g of acetic acid are dissolved...

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`.

A

0.0856

B

0.0956

C

0.1056

D

0.1156

Text Solution

Verified by Experts

The correct Answer is:
C
Wt . of `CH_3COOH` dissolved = 5g
Eq. of `CH_3COOH` dissolved = `5/60`
Volume of ethanol = 1 litre = 1000ml.
`:.` Weight of ethanol = 1000 × 0.789 = 789g
`:. "Molality of solution" = ("Moles of solute")/("wt of solvent in kg")`
`=(5/(60xx789))/1000=0.1056`
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