Question

18 gram glucose (Molar mass= 180) is dissolved in 100 ml of water at 300 K. If R = 30.0821 L-atm mol^-1 K^-1 what is the osmotic pressure of solution?

पर्याय

• 2.463 atm

• 24.63 atm

• 8.21 atm

• 0.821 atm

Answer 1

24.63 atm

Explanation:

The various quantities known to us are as follows:

R = 30.0821 L-atm mol^-1 K^-1 

w₂ = 18 gram

Molar mass (M₂) = 180

T = 300 K

V = 100 mL

To calculate the osmotic pressure of solution, we use the following formula,

`pi = ("w"_2"RT")/("M"_2"V")`

`= (18 xx 0.0821 xx 300 xx 1000)/(180 xx 100)`

= 24. 63 atm