What volume of 0.025 M MgCO3 can be made from 10.0 g of MgCO3?

The volume of 0.025 M MgCO3 that can be made from 10.0 g of MgCO3 is 4.76 L

calculation

Volume is calculated using Volume = moles / molarity formula

molarity = 0.025 M which is the same with 0.025 mol/L

moles = mass/molar mass

molar mass of MgCO3 = 24 + 12 + (16 x3) = 84 g/mol

moles is therefore = 10.0g / 84 g/mol = 0.119 moles

volume is therefore = 0.119 mol / 0.025 mol/L = 4.76 L