Mole fraction of solute in 4.5 molal aqueous solution is

Mole fraction, x = $\frac{\frac{\mathrm{w}}{\mathrm{m}}}{\frac{\mathrm{w}}{\mathrm{m}}+\frac{\mathrm{W}}{\mathrm{M}}}$

(where, w and m represent the mass and molar mass of solute and W and M represent the mass and molar mass of solvent.)

$\frac{1}{\mathrm{x}}=\frac{\frac{w}{m}+\frac{W}{M}}{{\displaystyle \frac{w}{m}}}$

$\frac{1}{\mathrm{x}}=1+\frac{W\times m}{M\times w}\times \frac{1000}{1000}$

$\frac{1}{\mathrm{x}}=1+\frac{1000}{M\times m}\left(\because Molality, m=\frac{w\times 1000}{\mathrm{m}\times \mathrm{W}}\right)$

$=1+\frac{1000}{81}$

$=\frac{1081}{81}=13.34$

$\mathrm{x}=\frac{1}{13.34}=0.075$