36 g of glucose (molar mass = 180) dissolved per litre of the solution has osmotic pressure of 4.98 bar at 27^oC. if the osmotic pressure of another solution is 1.52 bar at the same temperature, what will be its concentration? 

0.061 M

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)

\(\frac{1.52}{4.98} = \frac{C_2 \times 0.0821 \times 300}{0.2 \times 0.0831 \times 300}\)

\(⇒ \frac{1.52}{4.98} = \frac{C_2}{0.2}\)

\(⇒ C_2 = \frac{1.52 \times 0.2}{4.98}\)

\(⇒ C_2 = 0.061\, \ M\)