Determine weight of drug from molarity and volume of the solution

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

4.2.1.1) Molarity - general Normal 1


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How many grams of calcium gluconate (MW = 430 g/mol) should be dissolved in 500 mL of a solution so that the strength is 1.15 M?


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lb equals 247.25 g 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 500 mL, or 0.5 L.

    [1] Using formula method:
    `M=g/(aV) therefore g=MaV`
    `g=1.15 × 430 × 0.5 = 247.25` grams. Ans.

    [2] Using conceptual method:
    First, count the number of moles (`n`) of calcium gluconate present in this solution. 1.15 M means, there is 1.15 number of moles of calcium gluconate present in 1000 mL of the solution.Therefore,
    `(1.15\quad mol)/(1000 \quad mL)=x/(500 \quad mL) therefore x = 0.575 \quad mol`
    Then, calculate how many grams of calcium gluconate will be contained in 0.575 mol. Mole (`n`) is the gram weight (`g`) divided by molecular weight (`a`), or `n=g/a`. Therefore, `g = n × a = 0.575 \quad mol × (430 \quad g/(mol)) = 247.25` g. Ans.


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