Question

What is the molarity of a solution that contains 85.0 grams of Na2SO4 in 325 milliliters of solution? (The mass of one mole of Na2SO4 is 142 grams.)

0.195 M
0.599 M
1.84 M
6.22 M

Answer 1

first find the number of mol
85/142 = 0.598 mol

divide the mol by the volume in L
0.598/0.325 = 1.81M. C is close enough

Answer 2

Answer : The molarity of a solution is, 1.84 M

Explanation : Given,

Mass of
`Na_2SO_4` = 85.0 g

Molar mass of
`Na_2SO_4` = 142 g/mole

Volume of solution = 325 ml

Molarity : It is defined as the number of moles of solute present in one liter of solution.

In this question, the solute is
`Na_2SO_4`.

Formula used :

`Molarity=(w_b)/(M_b* V)* 1000`

where,

`w_b` = mass of solute
`Na_2SO_4`

`M_b` = molar mass of solute
`Na_2SO_4`

`V` = volume of solution in ml

Now put all the given values in the above formula, we get the molarity of the solution.

`Molarity=(85.0g)/(142g/mole* 325ml)* 1000=1.84mole/L=1.84M`

Therefore, the molarity of the solution is, 1.84 M