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

The vapour pressure of a dilute aqueous solution of glucose is 750 mm Hg at 373 K. Calculate:
(i) Molality
(ii) Mole fraction of the solute

A
(i) 0.013 m, (ii) 0.74
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(i) 0.74 m, (ii) 0.0132
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(i) 0.74 m, (ii) 0.98
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(i) 0.013 m, (ii) 0.26
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (i) 0.74 m, (ii) 0.0132
(ii) We know,
p0=Vapour pressure of water at 373 K = 760 mm Hg
Given,
Vapour pressure of solution = ps=750 mm
As the solution is dilute solution of glucose, hence
Molar mass of solvent MA=18 g mol1
Relative lowering of vapour pressure in dilute form:
ppsp=xB
Where, xB=mole fraction of solute
xB=760750760xB=0.0132
(i) Molality =xB×1000(1xB)×M
Where,
xB=mole fraction of soluteM=Molar mass of solvent=18 g mol1

Substituting the values, we get,
Molality=0.013×1000(10.013)×18 m
Molality=0.741 m

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon