$100 mL$ of $0.30 M$ glucose is mixed with $200 mL$ of $0.15 M$ $NaCl$ (complete dissociation, volumes additive) at $300 K$ . With $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , the osmotic pressure of the mixture is

Answer: B

Particles: glucose

0.100 \times 0.30 = 0.030

mol and

NaCl

0.200 \times 0.15 \times 2 = 0.060

mol, total

0.090

mol in

0.300

L, i.e.

0.300

π = 0.300 \times 0.0821 \times 300 = 7.39

atm. Ignoring dissociation gives 4.93 atm.