Determine volume of the solution from molarity and weight of drug

4. Moles and Molarity 4.2) Molarity 4.2.1) Molarity - general

4.2.1.1) Molarity - general Normal 1


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A pharmacists needs to prepare 1.8 M of calcium gluconate solution. If the quantity of calcium gluconate is 193.5 g, then determine the volume of the solution, in milliliter unit. The molecular weight of calcium gluconate is 430 g/mol.


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lb equals 250 mL kg

    Molarity is the number of moles (`n`) dissolved per liter of solution (`V`). Again, mole is the weight in gram (`g`) divided by the molecular weight (`a`).
    In this example, the volume of the solution is to be found out.

    [1] Using formula method:
    `M=g/(aV) therefore V=g/(Ma)=(193.5 \quad g)/((1.8 (mol)/L) × (430 g/(mol)))`
    `therefore V=0.25 \quad L` which is 250 mL. Ans.

    [2] Using conceptual method:
    First, count the number of moles (`n`) of calcium gluconate present in this solution. Mole is the gram amount of the solute divided by molecular weight. Therefore, `n= (193.5 \quad g)/(430 \quad g/(mol)) = 0.45 \quad mol`.
    Then, determination of the volume in which 0.45 mol of calcium gluconate will be present to give a molarity of 1.8. From the definition of molarity, 1.8 M means, 1.8 mole of solute will be dissolved in 1L of solution. Therefore, we can write,
    `(1.8 \quad mol)/(1 \quad L)=(0.45 \quad mol)/x therefore x = 0.25\quad L = 250\quad mL` Ans.