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.