36 g of glucose (molar mass = 180) dissolved per litre of the solution has osmotic pressure of 4.98 bar at 27oC. if the osmotic pressure of another solution is 1.52 bar at the same temperature, what will be its concentration?
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 0.061 M.
Osmotic pressure (\(\pi\)) is the pressure required to stop the osmotic flow of solvent through a semipermeable membrane. It is given by the formula:
\(\pi = C \cdot R \cdot T\)
where:
\(C\) is the concentration of the solute (in mol/L).
\(R\) is the ideal gas constant (\(0.0831 \text{ L·bar·K}^{-1}\text{·mol}^{-1}\)).
\(T\) is the temperature in Kelvin.
Given Data for the first solution
Osmotic Pressure (\(\pi_1\)) = \(4.98\, \ bar\)
Temperature (\(T\)) = \(27^oC = 300 K\) (since \(T = 27 + 273\))
Mass of glucose = \(36\, \ g\)
Molar Mass of glucose = \(180\, \ g/mol\)
\(\text{Number of moles} = \frac{\text{Mass of glucose}}{\text{Molar mass of glucose}} = \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol}\)
Since the volume of the solution is 1 L, thus concentration,
\(C_1 = \frac{\text{Number of moles}}{\text{Volume}} = \frac{0.2 \text{ mol}}{1 \text{ L}} = 0.2 \text{ M}\)
Thus, osmotic Pressure,
\(\pi_1 = C_1RT\)
\(⇒ 4.98 = 0.2 \times 0.0831 \times 300\) \(------(1)\)
Given Data for the second solution
Osmotic Pressure (\(\pi_2\))= 1.52 bar
Using Osmotic Pressure Formula:
\(\pi_2 = C_2 R T\)
\(⇒ 1.52 = C_2 \times 0.0821 \times 300\) \(-------(2)\)
Dividing equation (2) by equation (1)
\(\frac{1.52}{4.98} = \frac{C_2 \times 0.0821 \times 300}{0.2 \times 0.0831 \times 300}\)
\(⇒ \frac{1.52}{4.98} = \frac{C_2}{0.2}\)
\(⇒ C_2 = \frac{1.52 \times 0.2}{4.98}\)
\(⇒ C_2 = 0.061\, \ M\)