Question

\[10{\text{ g}}\] of cane sugar (molecular mass = 342) in \[1{\text{ }} \times {\text{ }}{10^{ - 3}}{m^3}\] of solution produces an osmotic pressure of \[6.68{\text{ }} \times {10^4}N{m^{ - 2}}\] 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 \[1{\text{ }} \times {\text{ }}{10^{ - 3}}{m^3}\] . Convert the unit of the volume into liters.
\[V = 1{\text{ }} \times {\text{ }}{10^{ - 3}}{m^3} \times \dfrac{{1000{\text{ L}}}}{{1{m^3}}} = 1{\text{ L}}\]
The osmotic pressure \[\pi \] is given as \[6.68{\text{ }} \times {10^4}N{m^{ - 2}}\] .
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
\[C = \dfrac{w}{{M.W \times V}} = \dfrac{{10{\text{ g}}}}{{342{\text{ g/mol }} \times {\text{ 1 L}}}} = 0.02924M\]
The van't Hoff factor ‘i’ is one as cane sugar is non electrolyte.
Write the expression for the osmotic pressure \[\pi \] 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.
\[
  \pi = i \times C \times R \times T \\
\Rightarrow 6.68{\text{ }} \times {10^4}N{m^{ - 2}} = 1 \times 0.02924M \times R \times 273{\text{ K}} \\
\Rightarrow R = 8368N{m^{ - 2}}L \\
\Rightarrow R = \;8.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}
 \]
The value of R is \[\;8.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}\]

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 \[1L = 0.001{m^3}\] and \[1{\text{ J = 1 Nm}}\]
This is as shown below.
\[
  R = 8368N{m^{ - 2}}L \times \dfrac{{0.001{m^3}}}{{1L}} \times \dfrac{{{\text{1 J}}}}{{{\text{1 Nm}}}} \\
\Rightarrow R = \;8.3684{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} \\
 \]