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 0.3 M of NaBr solution. If the quantity of NaBr is 23.1525 g, then determine the volume of the solution, in milliliter unit. The molecular weight of NaBr is 102.9 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)=(23.1525 \quad g)/((0.3 (mol)/L) × (102.9 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 NaBr present in this solution. Mole is the gram amount of the solute divided by molecular weight. Therefore, `n= (23.1525 \quad g)/(102.9 \quad g/(mol)) = 0.225 \quad mol`.
Then, determination of the volume in which 0.225 mol of NaBr will be present to give a molarity of 0.3. From the definition of molarity, 0.3 M means, 0.3 mole of solute will be dissolved in 1L of solution. Therefore, we can write,
`(0.3 \quad mol)/(1 \quad L)=(0.225 \quad mol)/x therefore x = 0.75\quad L = 750\quad mL` Ans.