Determine molarity from weight of drug 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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129 g of calcium gluconate is dissolved in 150 mL of solution. If the molecular weight of calcium gluconate is 430, then determine the molarity (M) of the solution


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lb equals 2 M 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 150 mL, or 0.15 L.

    [1] Using formula method:
    `M=g/(aV)=(129\quad g)/(430\quad (g/(mol)) × 0.15\quad L)`
    `therefore M = 2 (mol)/L=2` without unit, Ans.

    [2] Using conceptual method:
    First, count the number of moles (`n`) of calcium gluconate present in this solution:
    `=(weight\quad i n \quad g)/(mol ecu lar\quad wt)=(129\quad g)/(430 \quad (g/(mol)))=0.3 \quad mol`
    Then, calculate how many moles will be present in 1L of solution. Using equation, we can write:
    `(0.3\quad mol e)/(150 \quad mL)=x/(1000 \quad mL) therefore x = 2` Ans.