Question

A 5.05% (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 sulfate.....

Answer 1

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.

`\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\farc{100g}{1.03g/mL}=97.09mL`

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

Formula used :

`\text{Molarity of sodium sulfate solution}=\frac{\text{Mass of sodium sulfate}* 1000}{\text{Molar mass of sodium sulfate}* \text{Volume of solution (in mL)}}`

Molar mass of sodium sulfate = 142.04 g/mol

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

`\text{Molarity of sodium sulfate solution}=(5.05g* 1000)/(142.04g/mole* 97.09mL)=0.366mole/L`

Now we have to calculate the moles of barium sulfate.

`\text{Moles of }BaSO_4=\frac{\text{Mass of }BaSO_4}{\text{Molar mass of }BaSO_4}`

Molar mass of barium sulfate = 233.38 g/mol

`\text{Moles of }BaSO_4=(2.73g)/(233.38g/mol)=0.0117mol`

Now we have to calculate the moles of sodium sulfate.

The balanced chemical reaction will be:

`Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4`

From the balanced chemical reaction, we conclude that

As, 1 mole of
`BaSO_4` produced from 1 mole of
`Na_2SO_4`

So , 0.0117 mole of
`BaSO_4` produced from 0.0117 mole of
`Na_2SO_4`

Now we have to calculate the volume of sodium sulfate.

`Molarity=(Moles)/(Volume)`

`0.366mol/L=(0.0117mol)/(Volume)`

`Volume=0.0319L`

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