Question

Calculate molality of 2.5 g of ethanoic acid $\left[C{H}_{3}COOH\right]$ in 75 g of benzene.

• A. $0.551molk{g}^{-1}$
  
• B. $0.556molk{g}^{-1}$
  
• C. $0.558molk{g}^{-1}$
  
• D. $0.552molk{g}^{-1}$

Answer 1

The correct option is B $0.556molk{g}^{-1}$


MW of $C{H}_{3}COOH=60gmo{l}^{-1}$

$m=\frac{w_2\times 1000}{M{w}_2\times w_1}=\frac{2.5g\times 1000}{60gmo{l}^{-1}\times 75}=0.556molk{g}^{-1}$