### 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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