Question

Molality of $2.5\, g$ of ethanoic acid $ (CH_{3}COOH) $ in $75\, g$ of benzene is

• A. $ 0.565\,mol \,kg^{-1} $
• B. $ 0.656\, mol\, kg^{-1} $
• C. $ 0.556\,mol\,kg^{-1} $
• D. $ 0.665\,mol\,kg^{-1} $

Answer 1

Answer: (C)

Molality $( m )=\frac{\text { mass of solute }(\text { in } g ) \times 1000}{\text { molecular weight of solute } \times \text { massofsolvent }(\text { in } g )}$
$=\frac{2.5 \times 1000}{60 \times 7.5}=0.555\, mol\, kg ^{-1}$