Chemistry - Mole Concept Question with Solution | TestHub

The mole fraction of CH₃COOH ($X_{C{H}_{3}COOH}$) is 0.1.

This means $X_{H_{2}O}=1-0.1=0.9$.

A. $\%(w/w)=\frac{massofC{H}_{3}COOH}{totalmass}\times 100=\frac{6g}{6g+16.2g}\times 100=\frac{6}{22.2}\times 100\approx 27\%$. This statement is correct.

B. Assume 1 mole of solution. Moles of CH₃COOH = 0.1 mol. Moles of H₂O = 0.9 mol.

Mass of CH₃COOH = $0.1mol\times 60g/mol=6g$.

Mass of H₂O = $0.9mol\times 18g/mol=16.2g$.

Molality = $\frac{molesofsolute}{massofsolvent(kg)}=\frac{0.1mol}{0.0162kg}\approx 6.17m$. This statement is correct.

C. $\frac{N_{C{H}_{3}COOH}}{N_{H_{2}O}}=\frac{0.1}{0.9}=\frac{1}{9}$. So, $\frac{N_{H_{2}O}}{N_{C{H}_{3}COOH}}=9$. This statement is correct.

D. $\frac{W_{H_{2}O}}{W_{C{H}_{3}COOH}}=\frac{16.2g}{6g}=2.7$. This statement is incorrect.

The final answer is $\boxed{D}$