# Question #48ecb

Feb 18, 2016

${C}_{M} = \left({n}_{\text{total"))/(V_("total}}\right) = \frac{39.1 \times {10}^{- 4} m o l}{37.0 \times {10}^{- 3} L} = 0.106 M$

#### Explanation:

To find the concentration of chloride $C {l}^{-}$ in the mixture we will first find the total number of mole of $C {l}^{-}$ (${n}_{\text{total}}$) then divide it by the total volume (${V}_{\text{total}}$) of the mixture.

${C}_{M} = \left({n}_{\text{total"))/(V_("total}}\right)$

To find the ${n}_{\text{total}}$ we will calculate the number of mole coming from both salts:

${n}_{L i C l} = {C}_{M} \times V = 0.392 \frac{m o l}{\cancel{L}} \times 2.50 \times {10}^{- 3} \cancel{L} = 9.80 \times {10}^{- 4} m o l$

${n}_{K C l} = {C}_{M} \times V = 84.8 \times {10}^{- 3} \frac{m o l}{\cancel{L}} \times 34.5 \times {10}^{- 3} \cancel{L} = 29.3 \times {10}^{- 4} m o l$

${n}_{\text{total}} = {n}_{L i C l} + {n}_{K C l} = 9.80 \times {10}^{- 4} m o l + 29.3 \times {10}^{- 4} m o l = 39.1 \times {10}^{- 4} m o l$

The total volume is: ${V}_{\text{total}} = 34.5 \times {10}^{- 3} L + 2.50 \times {10}^{- 3} L = 37.0 \times {10}^{- 3} L$

The concentration can the be found by:

${C}_{M} = \left({n}_{\text{total"))/(V_("total}}\right) = \frac{39.1 \times {10}^{- 4} m o l}{37.0 \times {10}^{- 3} L} = 0.106 M$

Therefore, the correct choice is (c) 106 mM.