Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100 % ionized)

Question

Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100 % ionized) (A) 0.10 osmol (B) 0.20 osmol (C) 0.02 osmol (D) 0.004 osmol

Tardigrade Admin · Accepted Answer

Osmotic pressure is a colligative property, hence osmolarity is the molarity of the solution involving van’t Hoff factor

K_{4} [F e (C N_{6}] ⇌ y = 5 4 K^{+} + [F e (CN)_{6}]^{4 -}

i = [1 + (y - 1) x] = 1 + 4 x = 5 (x = 1)

Thus, osmolarity

= 0.02 \times i = 0.02 \times 5

= 0.10

osmol

Q. Osmolarity of 0.02M potassium ferrocyanide solution at 300K is (assume solute is 100% ionized)

0.10 osmol

0.20 osmol

0.02 osmol

0.004 osmol

Osmotic pressure is a colligative property, hence osmolarity is the molarity of the solution involving van’t Hoff factor K4​[Fe(CN6​]⇌y=54K++[Fe(CN)6​]4−​​ i=[1+(y−1)x]=1+4x=5(x=1) Thus, osmolarity =0.02×i=0.02×5 =0.10 osmol

Q. Osmolarity of $0.02 M$ potassium ferrocyanide solution at $300 K$ is (assume solute is $100%$ ionized)

Osmotic pressure is a colligative property, hence osmolarity is the molarity of the solution involving van’t Hoff factor
$K_{4} [F e (C N_{6}] ⇌ y = 5 4 K^{+} + [F e (CN)_{6}]^{4 -}$
$i = [1 + (y - 1) x] = 1 + 4 x = 5 (x = 1)$
Thus, osmolarity $= 0.02 \times i = 0.02 \times 5$
$= 0.10$ osmol