A 5 percent solutikon (by mass) of cane-sugar (M.W. 342) is isotonic with 0.877% solution of substnace X. find the molecular weight of X.

Explanation

The two solution are said to be isotonic if they have same osmotic pressure of cane sugar solution is equal to the osmotic pressure of ammonia solution. Let the osmotic pressure of cane sugar solution is `pi`, and osmotic pressure of Urea is `pi_(2)`
According to the quesion `pi_(1)=pi_(2)` ltBrgt `pi_(1)=w_(1)RT//M1V and pi_(2)=w_(2)RT//M_(2)w`
`W_(1)`=wt of cane sugar in 100 ml solution
`M_(1)`=mol. wt of can sugar.
`w_(2)`=wt of urea of 100 ml.
Solution `M_(2)=` mol. wt of urea.
`w_(1)=5,w_(2)=0.877,v=100ml`
`M_(1)` is given `=342,M_(2)=`
`n_(1)=n_(2)`
`w_(1)RT//m_(i)v=w_(2)RT//M_(2)V`
Putting the value of `w_(2)&w_(1)`.
`M_(2)=(w_(2))/(w_(1))M_(1)=0.877.342gram=60gram`