Question

What is the molearity of a solution that contains 85.0 grams of NaSO4 in 325ml of solution

Answer 1

`\huge{ \boxed{2.20 \: M}}`

Step-by-step explanation:

The molarity of the solution can be found by using the formula:

`c = (m)/(M * v)`

where:

c is the concentration in M , mol/dm³ or mol/L

v is the volume in L or dm³

m is the mass in grams

M is the molar mass in g/mol

From the question:

m = 85 g

Molar mass (M) of
`NaSO_4` = 23 + 32 + (16×4) = 23 + 32 + 64 = 119 g/mol

v = 325 mL

The volume has to be converted to Litres.

`if \: 1000ml = 1l \\ 325ml \: = (325)/(1000) * 1l \\ = 0.325 \: l`

`\therefore \: c = (85)/(119 * 0.325) \\ = 2.20 \: M`