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