Respuesta :
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.
[tex]\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\farc{100g}{1.03g/mL}=97.09mL[/tex]
Now we have to calculate the molarity of sodium sulfate solution.
Formula used :
[tex]\text{Molarity of sodium sulfate solution}=\frac{\text{Mass of sodium sulfate}\times 1000}{\text{Molar mass of sodium sulfate}\times \text{Volume of solution (in mL)}}[/tex]
Molar mass of sodium sulfate = 142.04 g/mol
Now put all the given values in this formula, we get:
[tex]\text{Molarity of sodium sulfate solution}=\frac{5.05g\times 1000}{142.04g/mole\times 97.09mL}=0.366mole/L[/tex]
Now we have to calculate the moles of barium sulfate.
[tex]\text{Moles of }BaSO_4=\frac{\text{Mass of }BaSO_4}{\text{Molar mass of }BaSO_4}[/tex]
Molar mass of barium sulfate = 233.38 g/mol
[tex]\text{Moles of }BaSO_4=\frac{2.73g}{233.38g/mol}=0.0117mol[/tex]
Now we have to calculate the moles of sodium sulfate.
The balanced chemical reaction will be:
[tex]Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4[/tex]
From the balanced chemical reaction, we conclude that
As, 1 mole of [tex]BaSO_4[/tex] produced from 1 mole of [tex]Na_2SO_4[/tex]
So , 0.0117 mole of [tex]BaSO_4[/tex] produced from 0.0117 mole of [tex]Na_2SO_4[/tex]
Now we have to calculate the volume of sodium sulfate.
[tex]Molarity=\frac{Moles}{Volume}[/tex]
[tex]0.366mol/L=\frac{0.0117mol}{Volume}[/tex]
[tex]Volume=0.0319L[/tex]
Thus, the volume of the sodium sulfate solution required is, 0.0319 L
