2M of 100 ml ${Na}_{2} {SO}_{4}$ is mixed with 3M of 100ml $N a C l$ solution and 1M of 200 ml ${CaCl}_{2}$ solution. Then the ratio of the concentration of cation and anion.

1/2

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1.5

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Solution

The correct option is D 1
Concentration of cations $= \frac{M_{1} V_{1} + M_{2} V_{2} + . . . . . . .}{V_{1} + V_{2} + . . . . . . .} = \frac{2 \times 100 \times 2 + 3 \times 100 \times 1 + 1 \times 200 \times 1}{400} = \frac{9}{4} M$
Concentration of anions $= \frac{2 \times 100 \times 1 + 3 \times 100 \times 1 + 1 \times 200 \times 2}{400} = \frac{9}{4} M$
Ratio $= \frac{9}{4} : \frac{9}{4}$
$⟹ 1 : 1$

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

The correct option is D 1Concentration of cations =M1V1+M2V2+.......V1+V2+.......=2×100×2+3×100×1+1×200×1400=94MConcentration of anions=2×100×1+3×100×1+1×200×2400=94MRatio =94:94⟹1:1

2M of 100 ml ${Na}_{2} {SO}_{4}$ is mixed with 3M of 100ml $N a C l$ solution and 1M of 200 ml ${CaCl}_{2}$ solution. Then the ratio of the concentration of cation and anion.