JEE Main 2023 — Solutions Question with Solution

To solve this problem, we will first calculate the osmotic pressure of the 0.05 M glucose solution and then use that information to determine the molecular mass of compound A.

1. Osmotic pressure equation:

$$\Pi =iMRT$$

where $$\Pi$$ is the osmotic pressure, $$i$$ is the van't Hoff factor (which is 1 for non-electrolytes), $$M$$ is the molarity, $$R$$ is the ideal gas constant (0.0821 L atm/mol K), and $$T$$ is the temperature in Kelvin.

Since both solutions have the same osmotic pressure at the same temperature, we can set their osmotic pressures equal to each other:

$${\Pi }_A={\Pi }_{glucose}$$

2. Calculate the osmotic pressure of the 0.05 M glucose solution:

Glucose is a non-electrolyte, so its van't Hoff factor is 1. We don't know the temperature, but since both solutions are at the same temperature, it will cancel out in our calculations.

$${\Pi }_{glucose}=(1)(0.05\mathrm{M})(R)(T)$$

3. Calculate the molarity of compound A:

Since we know that 12 g of compound A is dissolved in 1000 mL of water, we can find the molarity once we know the molecular mass.

Let $$x$$ be the molecular mass of compound A. Then, the molarity of compound A is:

$$M_A=\frac{12\mathrm{g}}{xg/mol}\times \frac{1}{1\mathrm{L}}=\frac{12}{x}\mathrm{M}$$

4. Set the osmotic pressures equal to each other:

$${\Pi }_A={\Pi }_{glucose}$$

$$(1)(\frac{12}{x}\mathrm{M})(R)(T)=(1)(0.05\mathrm{M})(R)(T)$$

The van't Hoff factors, ideal gas constant, and temperature cancel out:

$$\frac{12}{x}=0.05$$

5. Solve for the molecular mass (x) of compound A:

$$x=\frac{12}{0.05}$$

$$x=240g/mol$$

The molecular mass of compound A is 240 g/mol.