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

A 5% (w/V) solution of cane sugar is isotonic with 0.878% (w/V) solution of urea. Calculate the molecular mass of urea if the molecular mass of cane sugar is 342 g mol−1.

A
60.05 g mol1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
120.1 g mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
45.5 g mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
76.1 g mol1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 60.05 g mol1
Isotonic solution:
Two solutions having the same osmotic pressure across a semipermeable membrane is referred to as an isotonic solution. It has the same osmolarity (solute concentration), as another solution.
5% (w/V) cane sugar solution means, 5 g of cane sugar is there in 100 mL of solution
0.878% (w/V) urea solution means, 0.878 g of urea is there in 100 mL of solution
Molar concentration of cane sugar=w1m1×V1=5342×0.1
and
Molar concentration of urea=w2m2V2=0.878m2×0.1
For isotonic solutions,

w1m1V1=w2m2V25342×0.1=0.878m2×0.1m2=0.878×3425=60.05 g mol1

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