Which has maximum osmotic pressure at temperature T K?

Question

Which has maximum osmotic pressure at temperature T K? (A) 150 mL of 1.0 M urea solution (B) 210 mL of 1.0 M glucose solution (C) Mixture of 150 mL of 1.0 M urea solution and 210 mL of 1.0 M glucose solution (D) All of the above are isotonic solutions

Tardigrade Admin · Accepted Answer

π=MRT (a) π (urea) = RT (b) π (glucose) = RT (c) M (mixture) = (M1V1+M2V2/V1+V2) =(150×1+210×1/150+210)=1M ∴ π (mixture) = RT Thus, equal osmotic pressure. Thus, isotonic

Q. Which has maximum osmotic pressure at temperature T K?

150 mL of 1.0 M urea solution

210 mL of 1.0 M glucose solution

Mixture of 150 mL of 1.0 M urea solution and 210 mL of 1.0 M glucose solution

All of the above are isotonic solutions

$π = MRT$
(a) $π$ (urea) = RT
(b) $π$ (glucose) = RT
(c) M (mixture) $= \frac{M _{1} V _{1} + M _{2} V _{2}}{V _{1} + V _{2}}$
$= \frac{150 \times 1 + 210 \times 1}{150 + 210} = 1 M$
$∴ π$ (mixture) = RT
Thus, equal osmotic pressure. Thus, isotonic

Q. Which has maximum osmotic pressure at temperature T K?

150 mL of 1.0 M urea solution

210 mL of 1.0 M glucose solution

Mixture of 150 mL of 1.0 M urea solution and 210 mL of 1.0 M glucose solution

All of the above are isotonic solutions

π=MRT (a) π (urea) = RT (b) π (glucose) = RT (c) M (mixture) =V1​+V2​M1​V1​+M2​V2​​ =150+210150×1+210×1​=1M ∴π (mixture) = RT Thus, equal osmotic pressure. Thus, isotonic

$π = MRT$
(a) $π$ (urea) = RT
(b) $π$ (glucose) = RT
(c) M (mixture) $= \frac{M _{1} V _{1} + M _{2} V _{2}}{V _{1} + V _{2}}$
$= \frac{150 \times 1 + 210 \times 1}{150 + 210} = 1 M$
$∴ π$ (mixture) = RT
Thus, equal osmotic pressure. Thus, isotonic