1.5 % Urea solution and 5.25% unknown substance's solutions are isotonic. The density of both the solutions are `1 gm//cm^(3)` then what is the molecular weight of unknown substance ?

Options

A. 90.0 gram/mole
B. 115.0 gram/mole
C. 1050.0 gram/mole
D. 210.0 gram/mole

Answer

D

Explanation

For isotonic solution `(W_("unknown"))/(M_("unknown"))=(W_("urea"))/(M_("urea"))`
`M_("unknown")=(5.25xx60)/(1.5)=210` gram/mole