Question

If a $6.84\%\left(\frac{\mathrm{wt}}{\mathrm{vol}}\right)$ solution of cane sugar $(\mathrm{mol}.\mathrm{wt}=342)$ is isotonic with $1.52\%\left(\frac{\mathrm{wt}}{\mathrm{vol}}\right)$ solution of thiocarbamide, then the molecular weight of thiocarbamide is

Answer 1

Step 1: Given data

• Weight of cane sugar (${\mathrm{W}}_1$) = $6.84\mathrm{g}$
• Volume of cane sugar solution (${\mathrm{V}}_1$) = $100\mathrm{mL}$
• Molecular weight of cane sugar (${\mathrm{M}}_1$) = $342\mathrm{g}{\mathrm{mol}}^{-1}$
• Weight of Thiocarbamide (${\mathrm{W}}_2$) = $1.52\mathrm{g}$
• Volume of solution of Thiocarbamide (${\mathrm{V}}_2$) = $100\mathrm{mL}$

Step 2: Finding relation between the two solution

It is given to us that both of the solutions are isotonic with each other, thus their osmotic pressure is equal.

$\therefore$The osmotic pressure of solution of cane sugar = osmotic pressure of solution of Thiocarbamide

$\because$$\mathrm{Osmotic}\mathrm{pressure}=\mathrm{molar}\mathrm{concentration}\mathrm{of}\mathrm{dissolved}\mathrm{species}\left(\mathrm{C}\right)\times \mathrm{ideal}\mathrm{gas}\mathrm{constant}\left(\mathrm{R}\right)\times \mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{solution}\left(\mathrm{T}\right)$

$\therefore$$$\begin{gathered} {\mathrm{C}}_1\times \mathrm{R}\times \mathrm{T}={\mathrm{C}}_2\times \mathrm{R}\times \mathrm{T} \\ \Rightarrow {\mathrm{C}}_1={\mathrm{C}}_2 \end{gathered}$$ [$\because$R and T are equal on both sides]

Here ${\mathrm{C}}_1$is the concentration of the sugar cane solution and ${\mathrm{C}}_2$is the concentration of the Thiocarbamide solution.

Step 3: Finding the molecular weight of Thiocarbamide solution

Let us assume the weight of the Thiocarbamide solution be ${\mathrm{M}}_2$

Now, putting it in the formula of

$\mathrm{C}=\frac{\mathrm{W}\times 1000}{\mathrm{M}\times \mathrm{V}}$

$$\begin{gathered} \therefore {\mathrm{C}}_1={\mathrm{C}}_2 \\ \Rightarrow \frac{{\mathrm{W}}_1\times 1000}{{\mathrm{M}}_1\times {\mathrm{V}}_1}=\frac{{\mathrm{W}}_2\times 1000}{{\mathrm{M}}_2\times {\mathrm{V}}_2} \\ \Rightarrow \frac{6.84\times 1000}{342\times 100}=\frac{1.52\times 1000}{{\mathrm{M}}_2\times 100} \\ \Rightarrow {\mathrm{M}}_2=\frac{1.52\times 1000\times 342\times 100}{6.84\times 1000\times 100} \\ \Rightarrow {\mathrm{M}}_2=\frac{1.52\times 342}{6.84} \\ \Rightarrow {\mathrm{M}}_2=1.52\times 50 \\ \Rightarrow {\mathrm{M}}_2=76\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Therefore, the molecular weight of Thiocarbamide is $76\mathrm{g}{\mathrm{mol}}^{-1}$.