1.645 g of K3[Fe(CN)6] (molar mass 329 g mol^-1) is dissolved in water to make 500 mL of solution. The salt is 90% dissociated (K3[Fe(CN)6] -> 3 K+ + [Fe(CN)6]^3-, i = 1 + (n-1)alpha). Find the osmotic pressure at 27 degC in atm to two decimal places (R = 0.0821 L atm mol^-1 K^-1).

JEE Advanced levelNumerical answerDifficulty 3 of 3so107

$1.645 g$ of $K X_{3} [Fe (CN) X_{6}]$ (molar mass $329 g mo l^{- 1}$ ) is dissolved in water to make $500 mL$ of solution. The salt is $90%$ dissociated ( $K X_{3} [Fe (CN) X_{6}] 3 K X^{+} + [Fe (CN) X_{6}] X^{3 -}$ , $i = 1 + (n - 1) α$ ). Find the osmotic pressure at $27^{\circ} C$ in $atm$ to two decimal places ( $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ ).

Numerical answer type, answer in atm, tolerance 0.01.

n = 4

ions, so

i = 1 + 3 \times 0.90 = 3.70

C = (1.645/329) /0.500 = 0.0100 mol L^{- 1}

π = i CRT = 3.70 \times 0.0100 \times 0.0821 \times 300 = 0.911 atm