Question

$$10\text{ g}$$ of cane sugar (molecular mass = 342) in $$1\times {10}^{-3}{m}^3$$ of solution produces an osmotic pressure of $$6.68\times {10}^{4}N{m}^{-2}$$ at $$273\text{ K}$$. Calculate the value of R in SI units.
A. $$8.3684\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}$$
B. $$9.3684\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}$$
C. $$7.3684\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}$$
D. $$5.36841\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}$$

Answer 1

Hint:First calculate the molarity of cane sugar and then use the formula for the osmotic pressure. Take care of units.

Complete answer:
The volume of the solution is $$1\times {10}^{-3}{m}^3$$ . Convert the unit of the volume into liters.
$$V=1\times {10}^{-3}{m}^3\times {\displaystyle \frac{1000\text{ L}}{1m^3}}=1\text{ L}$$
The osmotic pressure $$\pi$$ is given as $$6.68\times {10}^{4}N{m}^{-2}$$ .
The absolute temperature T is $$273\text{ K}$$ .
The weight w of cane sugar is $$10\text{ g}$$
The molecular weight M.W of cane sugar is 342.
First calculate the concentration C of cane sugar
$$C={\displaystyle \frac{w}{M.W\times V}}={\displaystyle \frac{10\text{ g}}{342\text{ g/mol }\times \text{ 1 L}}}=0.02924M$$
The van't Hoff factor ‘i’ is one as cane sugar is non electrolyte.
Write the expression for the osmotic pressure $$\pi$$ of solution
$$\pi =i\times C\times R\times T$$
Substitute values in the above expression and calculate the value of the ideal gas constant R.
$$\begin{gathered} \pi =i\times C\times R\times T \\ \Rightarrow 6.68\times {10}^{4}N{m}^{-2}=1\times 0.02924M\times R\times 273\text{ K} \\ \Rightarrow R=8368N{m}^{-2}L \\ \Rightarrow R=8.3684\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1} \end{gathered}$$
The value of R is $$8.3684\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}$$

Hence, the correct option is the option A

Note:
You can convert the unit of R from $$8368N{m}^{-2}L$$ to $$8.3684\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}$$ by using two relations $$1L=0.001m^3$$ and $$1\text{ J = 1 Nm}$$
This is as shown below.
$$\begin{gathered} R=8368N{m}^{-2}L\times {\displaystyle \frac{0.001m^3}{1L}}\times {\displaystyle \frac{\text{1 J}}{\text{1 Nm}}} \\ \Rightarrow R=8.3684\text{ J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1} \end{gathered}$$