c
\(m=\frac{X_{\text {solute}}}{X_{\text {solvent}}} \times \frac{1000}{M_{\text {solvent}}}\)

\(m=\) molality, \(X_{\text {solute }}=\) Mole fraction of solute, \(X_{\text {solvent }}=\) Mole fraction of solvent, \(M_{\text {solvent }}=\) Molar mass of solvent

Here solute is methyl alcohol \((C{H_3}OH)\) and solvent is water \((H_{2} \mathrm { O } )\) Here water is solvent because question says solution is aqueous

Assume mole fraction of methyl alcohol \(=x\)

So mole fraction of water \(=1-X\)

and given molality \((\mathrm{m})=5.2\)

\(\mathrm{M}_{\text {solvent }}=18\) (as molar mass of \(\mathrm{H}_{2} \mathrm{O}=18\) )

\(\therefore 5.2=\frac{X}{1-X} \times \frac{1000}{18}\)

by solving this we get,

\(X=0.086\)