A $50 mL$ portion of a sol contains $0.35 M$ $NaCl$ as impurity. It is dialysed twice, each time against $200 mL$ of fresh pure water, allowing the salt to equilibrate completely between sol and water, and the water is discarded after each step. Find the final $NaCl$ concentration in the sol in mM, to the nearest integer.

Numerical answer type, answer in mM, tolerance 0.5.

Answer: 14 mM

After each equilibration the salt spreads over

50 + 200 = 250

mL, so the concentration is multiplied by

50/250 = 1/5

(the sol volume is taken as unchanged). After two steps

c = 0.35 \times (1/5)^{2} = 0.014 M

. This equals 14 mM.