Question

How many liters of a 3.86 M K2SO4 solution are needed to provide 72.9 g of K2SO4 (molar mass 174.01 g/mol)? Recall that M is equivalent to mol/L.

Answer 1

0.109 L of
`K_(2) S O_(4)` will be needed

Step-by-step explanation:

Given the molarity of
`K_(2) S O_(4)` is 3.86M

Molar mass of
`K_(2) S O_(4)` = 174.01 g/mol

Weight/Mass of solute
`K_(2) S O_(4)`is given as 72.9 g

Moles of
`K_(2) S O_(4)` based on the values given

=
`(72.9)/(174.01)=0.419 \mathrm{mol}`

By definition,

Molarity =
`\frac{\text { Moles of solute }}{\text {volume of solution}}`

And thus, volume =
`\frac{\text {Moles of solute}}{\text {Molarity}}`

Substituting the values in the above formula,

Volume =
`(0.419)/(3.86)=0.109 L`

Therefore, 0.109 L of
`K_(2) S O_(4)` will be needed