Molar mass of solute

Question

The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is:

(a) 5.8g mol^-1  (b) 0.58g mol^-1  (c) 58g mol^-1  (d) 0.88 g mol^-1         

Answer: (c) The elevation (ΔTb) in the boiling point
= 354.11 K - 353.23 K = 0.88 K
         
Substituting these values in expression
M_solute    = (K_b x 1000 x w)/ΔTb x W     
Where, w = weight of solute, W = weight of solvent
         
M_solute = (2.53 x 1.8 x 1000)/(0.88 x 90)  = 58 gm mol^-1
Hence, molar mass of the solute = 58 gm mol^-1