### 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

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.

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