Question

0 .15 g of a substance dissolved in 15 g of a solvent higher by 0.216$^\circ$C than that of the pure solvent. Find out the molecular weight of the substance. $(K_b\, for\, solvent \, is\, 2.16 ^\circ C)$

• A. 1.01
• B. 10.1
• C. 100
• D. 10

Answer 1

Answer: (C)

Elevation in boiling point, $\Delta T = K _{ b } Xm _{ b }$
Here, $K _{ b }=2.16, \Delta T =.216$
$m _{ b }=\Delta T / K _{ b }=>.216 / 2.16=0.1$
Now, molality, $m_{b}=$ Moles of solute/Mass of solvent(in kg)
$0.1 =$ Moles of solute/ $0.015$
$ \Rightarrow $ Moles of solute $=0.0015$
Finally, Moles= Mass (in gm)/ Mol. Wt
$0.0015=0.15$ / Mol wt
Mol w t $=100$