0.85%, aqueous solution of NaNO3 is apparently 90% dissociated at 27°C. The osmotic pressure will be (R = 0.082 atm K-1 mol-1)

Question

0.85%, aqueous solution of NaNO3 is apparently 90% dissociated at 27°C. The osmotic pressure will be (R = 0.082 atm K-1 mol-1) (A) 2.210 atm (B) 4.674 atm (C) 3.049 atm (D) 5.012 atm

Tardigrade Admin · Accepted Answer

Molecular weight of NaNO3 = 85 g mol-1 Molarity = (W×1000/M× V) = (0.85×1000/85×100) = 0.1 mol L-1 NaNO3 solution is 90 % dissociated beginmatrixNaNO3&leftharpoons&Na+&+&NO-3 1-0.9&&0.9&&0.9 endmatrix vanât Hoff factor, i = 1 - 0.9 + 0.9 + 0.9 = 1.9 ∴ π = 1.9 Ã 0.1 Ã 0.082 Ã 300 = 4.674 atm

Q. 0.85%, aqueous solution of NaNO3​ is apparently 90% dissociated at 27∘C. The osmotic pressure will be (R=0.082atmK−1mol−1)

2.210 atm

4.674 atm

3.049 atm

5.012 atm

Molecular weight of NaNO3​=85gmol−1 Molarity =M×VW×1000​=85×1000.85×1000​=0.1molL−1 NaNO3​ solution is 90% dissociated NaNO3​1−0.9​⇌​Na+0.9​+​NO3−​0.9​ van’t Hoff factor, i=1−0.9+0.9+0.9=1.9 ∴π=1.9×0.1×0.082×300=4.674atm

Q. 0.85%, aqueous solution of $N a N O_{3}$ is apparently 90% dissociated at $2 7^{\circ} C$ . The osmotic pressure will be $(R = 0.082 a t m K^{- 1} m o l^{- 1})$