If 0.44 gm of substance dissolved in 22.2 gm of benzene
lowers the freezing point of benzene by 0.567ºC, then the
molecular mass of substance is, the molal depression
constant= 5.12ºC mol⁻¹.
(a) 128.4 (b) 156.6 (c) 178.9 (d) 232.4
Answers
answer : option (c) 178.9
given, mass of substance = 0.44 gm
mass of benzene = 22.2 gm
molality, m = number of mole of solute/mass of solvent in Kg
= (0.44/M)/(22.2/1000) [ where M is molecular mass of substance ]
= 440/(22.2M) mol/Kg .........(1)
using formula, ∆T_f = k_f × m
⇒0.567°C = 5.12 °C/mol × 440/(22.2M) mol/Kg
⇒0.567 = 5.12 × 440/(22.2M)
⇒M = 5.12 × 440/(22.2 × 0.567)
= 178.9 g/mol
hence, option (c) is correct choice.
also read similar questions : 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40K. The free...
https://brainly.in/question/7070907
1.0g of a non electrolyte solute(molar mass 250g mol) was dissolved in 51.2 g of benzene. If the kf for benzene is 5.12k...
https://brainly.in/question/7898105