Question

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\frac{g}{m L}. \, $

• A. $0.1056$
• B. $0.056$
• C. $0.156$
• D. $0.16$

Answer 1

Answer: (A)

Wt. of $CH_{3}COOH$ dissolved $= \, 5 \, g$
Eq. of $CH_{3}COOH$ dissolved $=\frac{5}{60}$
Volume of ethanol $= \, 1 \, litre \, = \, 1000 \, mL$
$\therefore \, $ Weight of ethanol $= \, \left(\right.1000 \, \times \, 0.789\left.\right) \, g \, = \, 789 \, g \, $
$\therefore \, \, Molality \, of \, solution \, =\frac{M o l e s \, o f \, s o l u t e}{w t . \, o f \, s o l v e n t \, i n \, k g}$
$=\left\{\frac{5}{\frac{60 \times 789}{1000}}\right\}=0.1056$