Answer:
0.109 L of [tex]K_{2} S O_{4}[/tex] will be needed
Explanation:
Given the molarity of [tex]K_{2} S O_{4}[/tex] is 3.86M
Molar mass of [tex]K_{2} S O_{4}[/tex] = 174.01 g/mol
Weight/Mass of solute [tex]K_{2} S O_{4}[/tex]is given as 72.9 g
Moles of [tex]K_{2} S O_{4}[/tex] based on the values given
= [tex]\frac{72.9}{174.01}=0.419 \mathrm{mol}[/tex]
By definition,
Molarity = [tex]\frac{\text { Moles of solute }}{\text {volume of solution}}[/tex]
And thus, volume = [tex]\frac{\text {Moles of solute}}{\text {Molarity}}[/tex]
Substituting the values in the above formula,
Volume = [tex]\frac{0.419}{3.86}=0.109 L[/tex]
Therefore, 0.109 L of [tex]K_{2} S O_{4}[/tex] will be needed
