The molar mass of a solute X in g `mol^(-1)`, if its 1% solution is isotonic with a 5% solution of cane sugar (molar mass = `342 g mol^(-)`), is

Options

A. 68.4
B. 34.2
C. 136.2
D. 171.2

Answer

(a)

Explanation

According to isotonicity,`(w_(1))/(v_(1))=(w_(2))/(v^(2)m_(2))`
`(1)/(m_(1))=(5)/(342)`     (because`(w_(1))/(v_(1))=1, (w_(2))/(v_(2))=5`)
Therefore `m_(1)= 68.4`