Solution of $12 \mathrm{~g}$ of non-electrolyte (A) prepared by dissolving it in $1000 \mathrm{~mL}$ of water exerts the same osmotic pressure as that of $0.05 ~\mathrm{M}$ glucose solution at the same temperature. The empirical formula of $\mathrm{A}$ is $\mathrm{CH}_{2} \mathrm{O}$. The molecular mass of $\mathrm{A}$ is __________ g. (Nearest integer)

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. <br/><br/> 1. Osmotic pressure equation: <br/><br/> $\Pi = iMRT$ <br/><br/> 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. <br/><br/> Since both solutions have the same osmotic pressure at the same temperature, we can set their osmotic pressures equal to each other: <br/><br/> $\Pi_{A} = \Pi_{glucose}$ <br/><br/> 2. Calculate the osmotic pressure of the 0.05 M glucose solution: <br/><br/> 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. <br/><br/> $\Pi_{glucose} = (1)(0.05 ~\mathrm{M})(R)(T)$ <br/><br/> 3. Calculate the molarity of compound A: <br/><br/> 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. <br/><br/> Let $x$ be the molecular mass of compound A. Then, the molarity of compound A is: <br/><br/> $$M_{A} = \frac{12 ~\mathrm{g}}{x ~\mathrm{g/mol}} \times \frac{1}{1 ~\mathrm{L}} = \frac{12}{x} ~\mathrm{M}$$ <br/><br/> 4. Set the osmotic pressures equal to each other: <br/><br/> $\Pi_{A} = \Pi_{glucose}$ <br/><br/> $(1)(\frac{12}{x} ~\mathrm{M})(R)(T) = (1)(0.05 ~\mathrm{M})(R)(T)$ <br/><br/> The van't Hoff factors, ideal gas constant, and temperature cancel out: <br/><br/> $\frac{12}{x} = 0.05$ <br/><br/> 5. Solve for the molecular mass (x) of compound A: <br/><br/> $x = \frac{12}{0.05}$ <br/><br/> $x = 240 ~\mathrm{g/mol}$ <br/><br/> The molecular mass of compound A is 240 g/mol.