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Question: \[10{\text{ g}}\] of cane sugar (molecular mass = 342) in \[1{\text{ }} \times {\text{ }}{10^{ - 3}}...

10 g10{\text{ g}} of cane sugar (molecular mass = 342) in 1 × 10−3m31{\text{ }} \times {\text{ }}{10^{ - 3}}{m^3} of solution produces an osmotic pressure of 6.68 ×104Nm−26.68{\text{ }} \times {10^4}N{m^{ - 2}} at 273 K273{\text{ K}}. Calculate the value of R in SI units.
A. 8.3684 JK−1mol−18.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}
B. 9.3684 JK−1mol−19.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}
C. 7.3684 JK−1mol−17.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}
D. 5.36841 JK−1mol−15.36841{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}

Explanation

Solution

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 × 10−3m31{\text{ }} \times {\text{ }}{10^{ - 3}}{m^3} . Convert the unit of the volume into liters.
V=1 × 10−3m3×1000 L1m3=1 LV = 1{\text{ }} \times {\text{ }}{10^{ - 3}}{m^3} \times \dfrac{{1000{\text{ L}}}}{{1{m^3}}} = 1{\text{ L}}
The osmotic pressure π\pi is given as 6.68 ×104Nm−26.68{\text{ }} \times {10^4}N{m^{ - 2}} .
The absolute temperature T is 273 K273{\text{ K}} .
The weight w of cane sugar is 10 g10{\text{ g}}
The molecular weight M.W of cane sugar is 342.
First calculate the concentration C of cane sugar
C=wM.W×V=10 g342 g/mol × 1 L=0.02924MC = \dfrac{w}{{M.W \times V}} = \dfrac{{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 π\pi of solution
π=i×C×R×T\pi = i \times C \times R \times T
Substitute values in the above expression and calculate the value of the ideal gas constant R.

π=i×C×R×T ⇒6.68 ×104Nm−2=1×0.02924M×R×273 K ⇒R=8368Nm−2L ⇒R=  8.3684 JK−1mol−1 \pi = i \times C \times R \times T \\\ \Rightarrow 6.68{\text{ }} \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}}

The value of R is   8.3684 JK−1mol−1\;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 8368Nm−2L8368N{m^{ - 2}}L to   8.3684 JK−1mol−1\;8.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} by using two relations 1L=0.001m31L = 0.001{m^3} and 1 J = 1 Nm1{\text{ J = 1 Nm}}
This is as shown below.

R=8368Nm−2L×0.001m31L×1 J1 Nm ⇒R=  8.3684 JK−1mol−1  R = 8368N{m^{ - 2}}L \times \dfrac{{0.001{m^3}}}{{1L}} \times \dfrac{{{\text{1 J}}}}{{{\text{1 Nm}}}} \\\ \Rightarrow R = \;8.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} \\\