Question

How many grams of MgCO₃ are needed to make 255 mL of a 1.25 mol/L solution.

Answer 1

Using the formula for Molarity :

Molarity = No of moles /Volume of solution (in litres)

Given : Molarity of solution = 1.25 mol/L

Volume of solution = 255 mL => 0.255 L

Molarity of MgCO3 is given by:

Molarity = no of moles of MgCO3 / volume of solution

Putting the values, we get,

1.25 mol/L = no of moles of MgCO3 / 0.255 L

=> No of moles of MgCO3 = 0.31875 moles

As, no of moles = mass/Molar mass

Thus, Mass of MgCO3 / Molar mass of MgCO3 = 0.31875 moles

=> Mass of MgCO3 * molar mass of MgCO3 = 0.31875 moles

where Molar mass of MgCO3 = 84.3139 g/mol

Thus, Mass of MgCO3 = 0.31875 mol * 84.3139 g/mol

=> Mass of MgCO3 = 26.875 g or 26.88 g (approx)

Hence, 26.875 g of MgCO3 are required.