Answer to Question #171403 in General Chemistry for Andrei

Question

When MgSO4 with molar mass of 120g/mole are required to make 100.0mL of a solution that is 0.532M? How many grams of the Magnesium Sulfate will be needed to achieved the required solution? 

Answer 1

Solution:

Molarity of MgSO₄ = Moles of MgSO₄ / Solution volume

Hence,

Moles of MgSO₄ = Molarity of MgSO₄ × Solution volume

Moles of MgSO₄ = 0.532 mol/L × 0.1 L = 0.0532 mol

Moles of MgSO₄ = Mass of MgSO₄ / Molar mass of MgSO₄

Mass of MgSO₄ = Moles of MgSO₄ × Molar mass of MgSO₄

The molar mass of MgSO₄ is 120 g mol^-1.

Hence,

Mass of MgSO₄ = 0.0532 mol × 120 g mol^-1 = 6.384 g

Mass of MgSO₄ = 6.384 g

OR:

(0.1 L) × (0.532 mol MgSO₄ / 1 L) × (120 g MgSO₄ / 1 mol MgSO₄) = 6.384 g MgSO₄

Answer: 6.384 grams of magnesium sulfate (MgSO₄) will be needed to achieved the required solution.