MHT CET · Chemistry · Solutions

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?

A 2.463 atm
B 24.63 atm
C 8.21 atm
D 0.821 atm

Verified Solution

Answer & Solution

Correct Answer

(B) 24.63 atm

Step-by-step Solution

Detailed explanation

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 = \frac{w_{2} R T}{M_{2} V} = \frac{18 \times 0.0821 \times 300 \times 1000}{180 \times 100}

= 24.63 a t m

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app

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

Step-by-step Solution

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?