Molarity calculation - Random types
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 1.35 M of potassium citrate monohydrated solution. If the quantity of potassium citrate monohydrated is 328.05 g, then determine the volume of the solution, in milliliter unit. The molecular weight of potassium citrate monohydrated is 324 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)=(328.05 \quad g)/((1.35 (mol)/L) × (324 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 potassium citrate monohydrated present in this solution. Mole is the gram amount of the solute divided by molecular weight. Therefore, `n= (328.05 \quad g)/(324 \quad g/(mol)) = 1.0125 \quad mol`.
Then, determination of the volume in which 1.0125 mol of potassium citrate monohydrated will be present to give a molarity of 1.35. From the definition of molarity, 1.35 M means, 1.35 mole of solute will be dissolved in 1L of solution. Therefore, we can write,
`(1.35 \quad mol)/(1 \quad L)=(1.0125 \quad mol)/x therefore x = 0.75\quad L = 750\quad mL` Ans.