Answer:
1.1724 g.
Explanation:
- Firstly we need to calculate the percentage of pure MgSO₄ in (MgSO₄.7H₂O).
- If we have 1.0 mol of MgSO₄.7H₂O, then we will have the the mass of its molecular mass.
The molecular mass of 1.0 mol (MgSO₄.7H₂O) = 246.4 g/mol.
The molecular mass of pure MgSO₄ = 120.366 g/mol.
The molecular mass of 7(H₂O) = 7(18.0 g/mol) = 126.0 g/mol.
∴ The mass % of pure MgSO₄ = [(mass of pure MgSO₄)/(mass of MgSO₄.7H₂O)] x 100 = [(120.366 g/mol)/(246.4 g/mol)] x 100 = 48.85%.
∴ the quantity of pure MgSO₄ = (mass of MgSO₄.7H₂O)(% of MgSO₄/100) = (2.4 g)(48.85/100) = 1.1724 g.
Solubility at 100 °C = 480 g / 100 mL of solution =: saturated solution
The proportion 240 g / 50 mL is equal to the saturated ratio 480 g / 100 mL
Then, 240 g of sugar in 50 mL of boiling water will make a saturated solution.
Answer: saturated
I read and said it's a weak base