$1.17 g$ of $NaCl$ (molar mass $58.5 g mo l^{- 1}$ ) is dissolved in water to make $500 mL$ of solution at $300 K$ . Assuming complete dissociation and $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , the osmotic pressure is

Answer: C

n = 1.17/58.5 = 0.020

mol and

C = 0.020/0.500 = 0.040

M. With

i = 2

π = i CRT = 2 \times 0.040 \times 0.0821 \times 300 = 1.97

atm. Taking

i = 1

gives only half this value.