Question

find out osmotic pressure of 12% w/v kcl solution at 27°c​

Answer 1

Answer:

0.679

Explanation:

Acc to question,

5%sloution= glucose

V=100ml =0.1L

π = CRT

π = n^2/V RT

π =WB/Mb × Rt/V

=5/180×0.0821×298/0.1

π =0.679

Answer 2

Answer: 39.6 atm

Explanation:

π = WRT/MV

W = 12 g

M = 74.5 g/mol

V = 100ml = 0.1 L

R = 0.0821

T = 27°C = 300K

∴ π =  $\frac{12 * 0.0821 * 300}{74.5 * 0.1}$

∴ π = 39.6atm