JEE MAIN - Chemistry (2023 - 13th April Morning Shift - No. 19)
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)
Answer
240
Explanation
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.
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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