### 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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129 g of calcium gluconate is dissolved in 150 mL of solution. If the molecular weight of calcium gluconate is 430, then determine the molarity (M) of the solution

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lb equals 2 M 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 150 mL, or 0.15 L.

[1] Using formula method:
M=g/(aV)=(129\quad g)/(430\quad (g/(mol)) × 0.15\quad L)
therefore M = 2 (mol)/L=2 without unit, Ans.

[2] Using conceptual method:
First, count the number of moles (n) of calcium gluconate present in this solution:
=(weight\quad i n \quad g)/(mol ecu lar\quad wt)=(129\quad g)/(430 \quad (g/(mol)))=0.3 \quad mol
Then, calculate how many moles will be present in 1L of solution. Using equation, we can write:
(0.3\quad mol e)/(150 \quad mL)=x/(1000 \quad mL) therefore x = 2 Ans.