QJEEM23Chffb5: JEE Main 2023 Chemistry - Class 12 JEEM23Chffb5 Solution: Solution of of non-electrolyte (A) prepared by dissolving it in …

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:

$M_{A} = \frac{12 ~\mathrm{g}}{x ~\mathrm{g/mol}} \times \frac{1}{1 ~\mathrm{L}} = \frac{12}{x} ~\mathrm{M}$

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:

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

$x = 240 ~\mathrm{g/mol}$

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