A5.05% (by mass) of aqueaus solution of sodium sulfate
isallowed to react with an excess of barium chloride. what volume
ofthe sodium sulfate solution (density= 1.03g/ml) is required
toproduce 2.73 g of barium

Answer : The volume  of the sodium sulfate solution required is, 0.0319 L

Explanation :

As we are given that 5.05 % (by mass) of aqueous solution of sodium sulfate that means, 5.05 grams of sodium sulfate present in 100 grams of solution.

Mass of sodium sulfate = 5.05 g

Mass of solution = 100 g

Density of solution = 1.06 g/mL

First we have to calculate the volume of solution.

Now we have  to calculate the molarity of sodium sulfate solution.

Formula used :

Molar mass of sodium sulfate = 142.04 g/mol

Now put all the given values in this formula, we get:

Now we have to calculate the moles of barium sulfate.

Molar mass of barium sulfate = 233.38 g/mol

Now we have to calculate the moles of sodium sulfate.

The balanced chemical reaction will be:

From the balanced chemical reaction, we conclude that

As, 1 mole of produced from 1 mole of

So , 0.0117 mole of produced from 0.0117 mole of

Now we have to calculate the volume of sodium sulfate.

Thus, the volume  of the sodium sulfate solution required is, 0.0319 L

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