Question

x g of a non-electrolytic compound (molar mass = 200) are dissolved in 1.0 L of 0.05 M NaCl aqueous solution. The osmotic pressure of this solution is found to be 4.92 atm at 27°C. Calculate the value of x. Assume complete dissociation of NaCl and ideal behaviour of the solution (R = 0.082 L atm K⁻¹ mol⁻¹).

Answer 1

For the non-electrolytic compound:

w = x g, molar mass = 200, V = 1 L, T = 27°C = 300 K, R = 0.0821 L atm K⁻¹ mol⁻¹

Osmotic pressure (π₁) = ?

∴ Moles of solute (n₁) = `x/200`

∵ πV = n RT

∴ `pi_1 = (n_1RT)/V`

= `(x/200 xx 0.0821 xx 300)/1`

= 0.1231 x tm

For NaCl solution:

	NaCl        ⇌      Na+    +    Cl−
Initially	1 mole                   -               -
At equilibrium	-                      1 mole         1 mole

(as the dissociation is complete)

∴ `i = "No. of moles in solution"/"No. of moles added"`

= `(1 + 1)/1

= 2

Molarity of NaCl solution = 0.05 M

∴ Moles of NaCl (n₂) = 0.05, V = 1 L, T = 300 K, osmotic pressure (π₂) = ?

According to the modified equation of osmotic pressure

πV = i n RT

∴ π₂ × 1 = 2 × 0.05 × 0.0821 × 300

or π₂ = 2.463 atm

Given that the osmotic pressure of the solution of a non-electrolytic compound in 0.05 M NaCl at 27°C is 4.92 atm. This must be equal to the sum of osmotic pressure due to the compound (π₁) and due to 0.05 M NaCl (π₂). Therefore,

π₁ + π₂ = 4.92

or 0.1231x + 2.463 = 4.92

or `x = (4.92 - 2.463)/0.1231`

x = 19.96 g