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26 December, 13:31

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

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  1. 26 December, 16:53
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    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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