Question

What mass of CuSO4 would be required to prepare 5. 0 L of a 0. 215 M solution

Answer 1

`\huge{ \boxed{ \boxed{172 \: g}}}`

Step-by-step explanation:

The mass required to prepare the
`CuSO_4` solution can be found by using the formula

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

where

m is the mass in grams

v is the volume in L

c is the concentration in mol/l

M is the molar mass of the compound in g/mol

Making mass the subject we have;

m = c × M × v

Molar mass (M) of
`CuSO_4` = 64 + 32 + (16×4) = 160 g/mol

Substituting the values into the formula we have;

`m = 0.215 * 160 * 5 \\ = 172`

We have the final answer as;

172 g