0.1 l of solution contains 0.62 g of CuSO4 at a temperature of 19 ° C. The osmotic pressure of this solution

0.1 l of solution contains 0.62 g of CuSO4 at a temperature of 19 ° C. The osmotic pressure of this solution at a given temperature is 162 * 10 ^ 3 Pa. Calculate the apparent degree of dissociation of the salt.

Given:
V solution (CuSO4) = 0.1 l
m (CuSO4) = 0.62 g
T = 19oC = 292K
π = 162103 Pa = 162.103 kPa

To find:
α -?

Decision:
1) n (CuSO4) = m / M = 0.62 / 160 = 0.0039 mol;
2) Cm (CuSO4) = n / V p-ra = 0.0039 / 0.1 = 0.039 M;
3) i = π / (Cm * R * T) = 162.103 / (0.039 * 8.314 * 292) = 1.7121;
4) α = (i – 1) / (N (CuSO4 ions) – 1) = (1.7121 – 1) / (2 – 1) = 0.712.

Answer: The apparent degree of dissociation of CuSO4 is 0.712.