The molar mass of lithium carbonate is equal to 73.9 g/mol. Answer the following questions. 1) What mass of lithium carbonate is required to

The molar mass of lithium carbonate is equal to 73.9 g/mol. Answer the following questions. 1) What mass of lithium carbonate is required to prepare 500.0 mL of a 0.250 M Li2CO3 solution

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  1. Answer:

    9.2375 grams of lithium carbonate is required to prepare 500.0 mL of a 0.250 M Li₂CO₃ solution.

    Explanation:

    Molarity is a measure of the concentration of a substance and is defined as the number of moles of solute that are dissolved in a given volume.

    The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

    [tex]molarity=\frac{number of moles of solute}{volume}[/tex]

    Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].

    In this case:

    • molarity= 0.250 M
    • number of moles of solute= ?
    • volume= 500 mL= 0.5 L (being 1000 mL= 1 L)

    Replacing in the definition of molarity:

    [tex]0.250 M=\frac{number of moles of solute}{0.5 L}[/tex]

    Solving

    0.250 M*0.5 L = number of moles of solute

    0.125 = number of moles of solute

    If the molar mass of lithium carbonate is equal to 73.9 g/mol, then the required mass of the compound can be calculated by:

    [tex]73.9 \frac{g}{mol}*0.125 moles= 9.2375 grams[/tex]

    9.2375 grams of lithium carbonate is required to prepare 500.0 mL of a 0.250 M Li₂CO₃ solution.

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