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At 25^(@)C, at 5% aqueous solution of gl...

At `25^(@)C`, at 5% aqueous solution of glucose (molecular weight = 180 g `mol^(-1)`) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute ?

A

60

B

80

C

72

D

63

Text Solution

Verified by Experts

The correct Answer is:
C
Since the two solutions are isotonic, they must have same concentrations in moles/litre.
For glucose solution, concentration
`= 5g//100 cm^(3)` (Given)
= 50 g/L
`= (50)/(180)` moles/L
(`because` Molar mass of glucose = 180 g `mol^(-1)`)
For unknown substance, concentration
= 2g/100 `cm^(3)` (given)
= 20 g/L `= (20)/(M)` mol/L
`therefore" "(50)/(180) = (20)/(M)`
or M = 72
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