18 gram glucose (Molar mass = 180) is dissolved in 100 mol of water at 300 K. If R = 3 0.0821 L - a t m m o l - 1 K - 1 what is the osmotic pressure of solution?

Question

18 gram glucose (Molar mass = 180) is dissolved in 100 mol of water at 300 K. If R = 3 0.0821 L - a t m m o l - 1 K - 1 what is the osmotic pressure of solution?, 2.463 atm, 24.63 atm, 8.21 atm, 0.821 atm

MARKS App · Accepted Answer

The various quantities known to us are as follows: R = 0.0821 L - a t m m o l - 1 K - 1 w 2 = 18 g r a m Molar mass M 2 = 180 T = 300 K V = 100 m L To calculate the osmotic pressure of solution, we use the following formula, π = w 2 R T M 2 V = 18 × 0.0821 × 300 × 1000 180 × 100 = 24.63 a t m

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