Question

Dissolving $120 \,g$ of a compound of (mol. wt. $60$) in $1000 \,g$ of water gave a solution of density $1.12 \,g/mL$. The molarity of the solution is :

• A. 1.00 M
• B. 2.00 M
• C. 2.50 M
• D. 4.00 M

Answer 1

The Correct Option is B

To find the molarity of the solution, we must first determine the number of moles of solute and the volume of the solution in liters.

1. The molecular weight of the compound is given as 60 \, \text{g/mol}.
2. We have 120 \, \text{g} of the compound, which is:
   
   \text{Moles of solute} = \frac{\text{Mass of solute}}{\text{Molecular weight}} = \frac{120 \, \text{g}}{60 \, \text{g/mol}} = 2 \, \text{moles}
3. The density of the solution is given as 1.12 \, \text{g/mL}. The total mass of the solution is the sum of the mass of the compound and water, 120 \, \text{g} + 1000 \, \text{g} = 1120 \, \text{g}.
4. Converting the mass of the solution to volume using density:
   
   \text{Volume of the solution} = \frac{\text{Mass of the solution}}{\text{Density}} = \frac{1120 \, \text{g}}{1.12 \, \text{g/mL}} = 1000 \, \text{mL}
5. Convert the volume from milliliters to liters:
   1000 \, \text{mL} = 1 \, \text{L}
6. The molarity (M) of the solution is calculated as:
   
   \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} = \frac{2 \, \text{moles}}{1 \, \text{L}} = 2.00 \, \text{M}

Thus, the molarity of the solution is 2.00 M.