Question

How many grams of Na2SO4 are required to make 0.30 L of 0.500 M Na2SO4?

Answer 1

Answer: 21.3 g of
`Na_2SO_4` are required to make 0.30 L of 0.500 M
`Na_2SO_4`.

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

`Molarity=(n)/(V_s)`

where,

n = moles of solute

`V_s` = volume of solution in L

moles of
`Na_2SO_4` =
`\frac{\text {given mass}}{\text {Molar mass}}=(xg)/(142g/mol)`

Now put all the given values in the formula of molality, we get

`0.500M=(xg)/(142g/mol* 0.30L)`

`x=21.3g`

Therefore, 21.3 g of
`Na_2SO_4` are required to make 0.30 L of 0.500 M
`Na_2SO_4`.