An aqueous solution (0.85%) of NaNO3 is apparently 90% dissociated at 27°C. The osmotic pressure of

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An aqueous solution (0.85%) of NaNO3 is apparently 90% dissociated at 27°C. The osmotic pressure of solution is

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Molarity of solution = $\frac{0.85/85}{100/1000} = 0.1\text{ M}$

$\pi_{\text{theo}} = CRT = 0.1 \times 0.0821 \times 300 = 2.463\text{ atm} $$${\text{i = 1 + }\alpha(n - 1) = 1 + 0.9(2 - 1) = 1.9 }{\therefore\pi_{\text{exp}} = i.\pi_{\text{theo}} = 1.9 \times 2.463 = 4.58\text{ atm}}$$

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