A solution containing 5.0 g urea per litre was found to be isotonic with 0.7 percent (wt./vol.) solution of an organic, non-volatile solute. Calculate the molar mass of the organic compound (molar mass of urea=60).

Explanation

Solution are isotonic, therefore `pi_(1)=pi_(2)`
`n_(1)/V_(1) ST=(n_(2))/V_(2) ST rArr n_(1)/V_(1)=n_(2)/V_(2)` (S and T are constants)
`(w_1)/(m_1 xx V_(t))_("urea")=(w_(2))/(m_(2) xx V_(2))_("organic") rArr (5.0)/(60 xx 1000)=(0.7)/(m_(2) xx 100)`
Molar mass of the unknown organic compound `m_(2)=(60 xx 1000 xx 0.7)/(5.0 xx 100)=84`