Determine molarity from weight of drug 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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36 g of potassium bicarbonate is dissolved in 400 mL of solution. If the molecular weight of potassium bicarbonate is 100, then determine the molarity (M) of the solution
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lb equals 0.9 M 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 400 mL, or 0.4 L.
[1] Using formula method:
`M=g/(aV)=(36\quad g)/(100\quad (g/(mol)) × 0.4\quad L)`
`therefore M = 0.9 (mol)/L=0.9` without unit, Ans.
[2] Using conceptual method:
First, count the number of moles (`n`) of potassium bicarbonate present in this solution:
`=(weight\quad i n \quad g)/(mol ecu lar\quad wt)=(36\quad g)/(100 \quad (g/(mol)))=0.36 \quad mol`
Then, calculate how many moles will be present in 1L of solution. Using equation, we can write:
`(0.36\quad mol e)/(400 \quad mL)=x/(1000 \quad mL) therefore x = 0.9` Ans.