Question

Determine the quantity (g) of pure MgSO4 in 2.4 g of MgSO4•7H2O. Show your work.

Answer 1

Answer: The quantity of pure
`MgSO_4` in 2.4 g of
`MgSO_4.7H_2O` is 1.17 g

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}=(2.4g)/(246g/mol)=0.0098moles`

As 1 mole of
`MgSO_4.7H_2O` contains = 1 mole of
`MgSO_4`

Thus 0.0098 moles of
`MgSO_4.7H_2O` contains =
`(1)/(1)* 0.0098=0.0098mole` of
`MgSO_4`

Mass of
`MgSO_4=0.0098mol* 120g/mol=1.17g`

Thus the quantity of pure
`MgSO_4` in 2.4 g of
`MgSO_4.7H_2O` is 1.17 g