Question

If 0.15 g of a solute, dissolved in 15 g of solvent, increases the boiling point by ${0.216}^{o}C$ over that of the pure solvent, the molecular mass of the substance is (Molal elevation constant for the solvent is 2.16^OC$):

• A. 1.01
• B. 10
• C. 10.1
• D. 100

Answer 1

The correct option is D 100

Elevation in boiling point, ∆T = K_b X m_b

Here, K_b = 2.16 , ∆T = .216

m_b= ∆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 => Moles of solute= 0.0015\

Finally, Moles= Mass (in gm)/ Mol. Wt

0.0015= 0.15/Mol wt

Mol wt= 100