Question

1.5 g of a non-volatile, non-electrolyte is dissolved in 50 g benzene ($K_b$ = 2.5 K kg mol⁻¹). The elevation of the boiling point of the solution is 0.75 K. The molecular weight of the solute in g mol⁻¹ is
(a) 200
(b) 50
(c) 75
(d) 100

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

(d) 100

Explanation:

ΔTb=Kb⋅m

where,

ΔTb is the boiling point elevation,

Tb (solution) − Tb (pure solvent),

Kb is the ebullioscopic constant,

m is the molality of the solution

ΔTb=0.75K

weight of solute =1.5g

weight of solvent =50g

molality =

weight of solute molar weight of solute 1000 weight of solvent in g=1.5M×100050g

Now,0.75K=2.5Kkgmol−1×1.5M×100050

M=2.5Kkgmol−1×1.5×10000.70K×50=100.8gmol−1

Molecular weight of solute is 100.8gmol−1.

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