$100 mL$ of $0.10 M$ $NaCl$ (assume complete dissociation) is mixed with $100 mL$ of $0.30 M$ glucose. Volumes are additive. The osmotic pressure of the mixture at $310 K$ ( $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ ) is

Answer: C

Moles of particles:

NaCl

gives

2 \times 0.0100 = 0.0200

and glucose gives

0.0300

, total

0.0500 mol

0.200 L

, i.e.

0.250 mol L^{- 1}

π = 0.250 \times 0.0821 \times 310 = 6.36 atm