What volume of $0.8 M A l C l_{3}$ solution should be mixed with $50$ mL of $0.2 M C a C l_{2}$ solution to get solution of chloride ion concentration equal to $0.6 M$ ?

5.56

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5.87

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5.90

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6.01

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Solution

The correct option is A $5.56$ mL
Let $V$ mL of $0.8$ M $A l C l_{3}$ be mixed with $50$ mL of $0.2$ M $C a C l_{2}$

Total number of millimoles $= 3 \times 0.8 \times V + 2 \times 0.2 \times 50 = 2.4 V + 20$

The total volume of the solution $= V + 50$ mL.

Molarity $= \frac{N u m b e r o f m o l e s o f c h l o r i d e i o n s}{V o l u m e o f t h e s o l u t i o n} = \frac{2.4 V + 20}{V + 50} = 0.6 M$ .

$2.4 V + 20 = 0.6 V + 30$

$1.8 V = 30 - 20$

$V = \frac{10}{1.8} = 5.56$ ml

Hence, option A is correct.

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What volume of 0.8MAlCl3 solution should be mixed with 50 mL of 0.2MCaCl2 solution to get solution of chloride ion concentration equal to 0.6 M?

What volume of $0.8 M A l C l_{3}$ solution should be mixed with $50$ mL of $0.2 M C a C l_{2}$ solution to get solution of chloride ion concentration equal to $0.6 M$ ?