### Determine volume of the solution from molarity and weight of drug

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 1.8 M of calcium gluconate solution. If the quantity of calcium gluconate is 193.5 g, then determine the volume of the solution, in milliliter unit. The molecular weight of calcium gluconate is 430 g/mol. Click on the button below to see the answer and explanations

lb equals 250 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.

 Using formula method:
M=g/(aV) therefore V=g/(Ma)=(193.5 \quad g)/((1.8 (mol)/L) × (430 g/(mol)))
therefore V=0.25 \quad L which is 250 mL. Ans.

 Using conceptual method:
First, count the number of moles (n) of calcium gluconate present in this solution. Mole is the gram amount of the solute divided by molecular weight. Therefore, n= (193.5 \quad g)/(430 \quad g/(mol)) = 0.45 \quad mol.
Then, determination of the volume in which 0.45 mol of calcium gluconate will be present to give a molarity of 1.8. From the definition of molarity, 1.8 M means, 1.8 mole of solute will be dissolved in 1L of solution. Therefore, we can write,
(1.8 \quad mol)/(1 \quad L)=(0.45 \quad mol)/x therefore x = 0.25\quad L = 250\quad mL Ans.