olecular 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.

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