Question

A solution containing 10g per $d{{m}^{3}}$ of urea (molecular mass = 60g$mo{{l}^{-1}}$) is isotonic with a 5% solution of a non-volatile solute. Calculate the molecular mass of this non-volatile solute.

Answer 1

Hint: Isotonic solutions contain equal concentrations of impermeable solutes on either side of the membrane and so the cell neither swells or shrinks.

Complete step-by-step answer:
In brief we know that isotonic solutions have the same concentration i.e. the same molarity. It is given that the solution contains 10g per $d{{m}^{3}}$urea.
And we know that the chemical formula of urea is $C{{H}_{4}}{{N}_{2}}O$ .
So, molecular mass of urea = 12 + 4 x (1) + 2 x (14) +16
= 60g$mo{{l}^{-1}}$.
So, the molarity of urea = mass concentration / molar mass = $(\dfrac{10g/d{{m}^{3}}}{60})=\dfrac{1}{6}mol/d{{m}^{3}}$
Let’s assume the molar mass of the non-volatile solution is ‘m’.
So, the molarity of non- volatile solute =mass concentration / molar mass
= $\dfrac{50g/d{{m}^{3}}}{m}$.
Both solutions are isotonic with each other means concentration of both solutions are same:
Molarity of urea = Molarity of non-volatile solution
$(\dfrac{10g/d{{m}^{3}}}{60})=\dfrac{1}{6}mol/d{{m}^{3}}$= $\dfrac{50g/d{{m}^{3}}}{m}$
By solving the above equation we get the value of ‘m’ is:
$m=50\times 6gmo{{l}^{-1}}$
= 300$gmo{{l}^{-1}}$
So, the molecular mass of non-volatile solute is “300$gmo{{l}^{-1}}$”

Note: In between these two solutions, solution osmosis is not possible. So, their molar concentrations are equal to each other.