wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A solution containing 0.5216 g of non-volatile solute 'A' (molar wt. = 128.16 g mol1) in 50 mL of CCl4 shows boiling-point elevation of 0.402oC while a solution of 0.6216 g of an unknown solute in the same weight of solvent gave a boiling-point elevation of 0.647oC. Find the molecular mass of the unknown solute.

A
130 g/mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
99.5 g/mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
85.5 g/mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
94.84 g/mol
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 94.84 g/mol
Boiling point elevation of a solution is given by,
Tb=Kb×m
where,
Tb is the elevation in boiling point.
Kb is molal elevation constant.
m is molality of solution.

Mass of the solute 'A' in first solution =0.5216 g
No. of moles of solute 'A' in first solution =0.5216128.16
Suppose the weight of 50 mL of solvent CCl4 is W g.
For the first solution,
Molality =0.5216/128.16W×1000
=521.6128.16 W
Kb=Tbm
=0.402521.6/128.16 W
As for the second solution,
Molality =0.6216/MW×1000
=621.6WM
Kb=0.647621.6/WM
(M is the molar weight of the unknown solute in second solution.)

Since, Kb is the property of the solvent and is independent of the solute.
For the same solvent, Kb=Kb
Hence, 0.402521.6/128.16 W=0.647621.6/WM
M=94.84 g/mol

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon