Chemistry - Liquid Solution Question with Solution | TestHub

Here it is given that$\text{w}\text{=}\text{0}\text{.15}\text{g,}{\text{ΔT}}_{\text{b}}\text{=}\text{0}{\text{.216}}^{\text{o}}\text{C}$
$\text{W}\text{=}\text{15}\text{g,}{\text{K}}_{\text{b}}\text{=}\text{2}\text{.16}\text{Kkg}{\text{mol}}^{-1}$
Substituting values in the expression,$\mathrm{m}=\frac{1000\times {\mathrm{K}}_{\mathrm{b}}\times \mathrm{w}}{\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}\times \mathrm{W}}$
$\text{m}\text{=}\frac{\text{1000×216×0}\text{.15}}{\text{0}\text{.216×15}}\text{=100}\text{g/mol}$