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
