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

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