A $0.040 M$ solution of $K X_{3} [Fe (CN) X_{6}]$ is $70%$ dissociated ( $K X_{3} [Fe (CN) X_{6}] 3 K X^{+} + [Fe (CN) X_{6}] X^{3 -}$ ; use $i = 1 + (n - 1) α$ ). At the same temperature, what mass of glucose (molar mass $180 g mo l^{- 1}$ ) per litre of solution is isotonic with it? Give the answer in g to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

Answer: 22.32 g

Each formula unit gives

n = 4

ions, so

i = 1 + 3 \times 0.70 = 3.10

. Effective concentration

= 0.040 \times 3.10 = 0.124

M. Isotonic glucose is

0.124

M, so mass per litre

= 0.124 \times 180 = 22.32