If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by `0.216 ^@C`, than that of the pure solvent, the molecular weight of the solute. (Molal elevation constant for the solute is `2.16 ^@C` ) is

Options

A. 1.01
B. 10.1
C. 10
D. 100

Answer

D

Explanation

`M_A=(w_AxxK_bxx1000)/(w_B xx DeltaT_b)`
`w_A` = weight of solute = 0.15 g, `w_B` = weight of solvent = 15 g
`DeltaT_b = 0.126^@C`
`K_b = 2.16^@C, M_A`, = Mol. mass of solute
`:. M_A = (0.15 xx 2.16 xx 1000)/(15 xx 0.216) = 100 g`