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Ways of Representing the Concentration of a Solution

2M of 100ml...

Question

$2$ M of $100$ ml of $N a_{2} S O_{4}$ is mixed with $3$ M of $100$ ml $N a C l$ solution and $1$ M of $200$ ml $C a C l_{2}$ solution.
Then the ratio of the concentration of cation and anion.

1 / 2

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2

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1.5

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1

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Solution

The correct option is D $1$
$m o l e = \frac{M \times 1000}{V (i n m l)}$

$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{2 -}$
0.2 0 0
0.4 0.2
$[N a^{+}] = (0.4 / 100) \times 1000 = 4 M$

$[S O_{4}^{2 -}] = (0.2 / 100) \times 1000 = 2 M$
$N a C l \to N a^{+} + C l^{-}$
0.3 0 0
0.3 0.3

$[N a^{+}] = [C l^{-}] = (0.3 / 100) \times 1000 = 3 M$

$C a C l_{2} \to C a^{2 +} + 2 C l^{-}$
0.2 0 0
0.2 0.4

$[C a^{2 +}] = (0.2 / 100) \times 1000 = 2 M$

$[C l^{-}] = (0.4 / 100) \times 1000 = 4 M$
Ratio of concentration of cation and anion $= \frac{4 + 3 + 2}{2 + 3 + 4} = 1$
Option A is correct.

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2M of 100ml of Na2SO4 is mixed with 3M of 100ml NaCl solution and 1M of 200ml CaCl2 solution. Then the ratio of the concentration of cation and anion.

$2$ M of $100$ ml of $N a_{2} S O_{4}$ is mixed with $3$ M of $100$ ml $N a C l$ solution and $1$ M of $200$ ml $C a C l_{2}$ solution.
Then the ratio of the concentration of cation and anion.