What is the molarity of a solution containing 14.6 g of Ca(NO3)2 in 221 mL?

To calculate the molarity of a solution, you can use the formula:

\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]

Step 1: Calculate the moles of solute (in this case, \( Ca(NO_3)_2 \)). First, determine the molar mass of \( Ca(NO_3)_2 \). The atomic masses are approximately: Ca = 40.08 g/mol, N = 14.01 g/mol, O = 16.00 g/mol. Therefore, the molar mass of \( Ca(NO_3)_2 \) is: \[ (40.08) + 2 \times [ (14.01) + 3 \times (16.00) ] = 40.08 + 2 \times (14.01 + 48.00)= 40.08 + 2 \times 62.01 = 40.08 + 124.02 = 164.10 \text{ g/mol} \]

Step 2: Use the molar mass to find the number of moles of \( Ca(NO_3)_2 \) in 14.6 g. \[ \text{Number of moles} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{14.6 \text{ g}}{164.10 \text{ g/mol}} \]

Step 3: Convert the volume of solution from milliliters to liters, because molarity is expressed in moles/liter. \[ 221 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.221 \text{ L} \]

Step 4: Now calculate the molarity: \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \] \[ \text{Molarity} = \frac{14.6 \text{ g}/164.10 \text{ g/mol}}{0.221 \text{ L}} \] \[ \text{Molarity} = \frac{0.0889 \text{ moles}}{0.221 \text{ L}} \] \[ \text{Molarity} = 0.402 \text{ M} \]

So, the molarity of the solution is approximately 0.402 M.