Answer : 17.12 g
Explanation:\(\Delta T =k_b\times m\)
\(\Delta T\) = elevation in boiling point
\(k_b\) = boiling point elevation constant
m= molality
\(molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}\)
given \(\Delta T=6.28^{\circ}C\)
Molar mass of solute = 46.0 \(gmol^{-1}\)
Weight of the solvent = 150.0 g = 0.15 kg
Putting in the values
\(molality=\frac{x}{46gmol^{-1}\times0.15kg}\)
\(6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}\)
x = 17.12 g
