$\Delta {T}_{b}={i} {K}_{{b}} {m}$

$\Delta {T}_{{b}}=\frac{1}{2} \times 2.6 \times \frac{1.22 / {M}_{{w}}}{100 / 1000}...(1)$

With Acetone as solvent

$\Delta {T}_{{b}}={i} {K}_{6} {~m}$

$0.17=1 \times 1.7 \times \frac{1.22 / {M}_{{w}}}{100 / 1000}...(2)$

$(1) /(2)$

$\frac{\Delta {T}_{{b}}}{0.17}=\frac{\frac{1}{2} \times 2.6+\frac{1.22 / {M}_{{w}}}{100 / 1000}}{1 \times 1.7 \times \frac{1.22 / {M}_{{w}}}{100 / 1000}}$

$\Delta {T}_{{b}}=\frac{0.26}{2}$

$\Delta {T}_{{b}}=13 \times 10^{-2}$

$\Rightarrow {x}=13$