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 0.55 M of calcium gluconate solution. If the quantity of calcium gluconate is 177.375 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 750 mL kg
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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)=(177.375 \quad g)/((0.55 (mol)/L) × (430 g/(mol)))`
`therefore V=0.75 \quad L` which is 750 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= (177.375 \quad g)/(430 \quad g/(mol)) = 0.4125 \quad mol`.
Then, determination of the volume in which 0.4125 mol of calcium gluconate will be present to give a molarity of 0.55. From the definition of molarity, 0.55 M means, 0.55 mole of solute will be dissolved in 1L of solution. Therefore, we can write,
`(0.55 \quad mol)/(1 \quad L)=(0.4125 \quad mol)/x therefore x = 0.75\quad L = 750\quad mL` Ans.