Question

33. What is osmotic pressure of the aq. solution of the
given solute at 27°C if depression in freezing point is
0.93°C ? Molal depression constant of water
1.86K mol kg ? Assume molality as molarity.​

Answer 1

$\huge\mathfrak\pink{\underline{\underline{AnSwer}}}$

★ As we know,

∆Tf = i × Kf × m

where,

Kf = molal depression constant or cryoscopic constant.

given,

Kf = 1.86K/molal

∆Tf = 0.93 ℃

now,

0.93 = 1 × 1.86 × m

=> m = 0.5

also,

π =i c R T

π= osmotic pressure

C = molarity = 0.5

R =0.821 L -atm

T = 273 + 27℃ = 300

putting these value in equation 1 we get,

π = 1 × 0.5 × 0.821 × 300

π = 12.315 atm

Happy to help !

:)