Question

Calculate M2. 0.822 g of solute in 300 mdm3

of H2O. Osmotic

pressure of solution is 149 m
mHg at 298 K​

Answer 1

Answer:

given = W2 = 0.822g

              R = 0.08205dm³atmK⁻¹mol⁻¹

              T = 298K

               V = 300dm³

               π = 149mmHg

to find = M2

solution =

Π = W1RT/M2V

Π= 149/760= 0.196atm ...(convert the given atmospheric pressure into atm )

M2 =0.822gX 0.08205dm³atmK⁻¹mol⁻¹X 298K/0.196atmX 0.3dm³

M2 =  0.0674451 X298K / 0.0588

M2 =  342gmol⁻¹

M2. 0.822 g of solute in 300 mdm3  of H2O. Osmotic  pressure of solution is 149 m mHg at 298 K​ is 342gmol⁻¹