Q. Elevation in boiling point was $0.52℃$ when 6 g of a compound was dissolved in 100 g of water. Molecular weight of X is ($k_{b}\ of\$water is $5.2℃$ per 100 g water )
A) 120
B) 60
C) 600
D) 180
Explanation
(b)
M=$\frac{100 \times k_{b} \times w}{\mathrm{\Delta}T_{b} \times W}$
Given, ${\ \ \ \ k}_{b} = 5.2$
w= mass of solute = 6 g
W=mass of solvent =100 g
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}T_{b} =$ elevation in boiling point = 0.52$℃$
M= molecular weight =?
$\therefore$ M =$\frac{100 \times 5.2 \times 6}{0.52 \times 100} = 60$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.