Molarity Calculation for Strontium Chloride Solution

Molarity Problem Sample

Objective: Prepare a 1 M (molar) solution of strontium chloride (SrCl₂).
Given: 5 g of strontium chloride available in a bottle.
Molar Mass Calculation: The formula weight of strontium chloride (SrCl₂) is given as 158.53 g/mol.

Molarity (M): Molarity is defined as the number of moles of solute per liter of solution. It can be expressed mathematically as: M = \frac{n}{V} where:
- M = molarity
- n = moles of solute
- V = volume of solution in liters

**Calculate Moles of Strontium Chloride:
- Use the formula to find the number of moles (n):**
  n = \frac{mass}{molar \ mass}
- For this problem:
  n = \frac{5\text{ g}}{158.53\text{ g/mol}}
- Calculate the moles:
  n \approx 0.0316 \text{ moles}
Determine Volume of Solution in Liters needed to create a 1 M Solution:
- Rearrange the molarity equation to find the volume (V):
  V = \frac{n}{M}
- Substitute the values:
  V = \frac{0.0316 \text{ moles}}{1\text{ M}}
- Calculate the volume:
  V \approx 0.0316 \text{ L}
Convert Volume to Milliliters:
- To convert liters to milliliters, multiply by 1000.
- (0.0316 ext{ L}) (1000 ext{ mL/L}) \approx 31.6 ext{ mL}

Therefore, the total volume of the resulting solution when using the entire 5 g of strontium chloride to prepare a 1 M solution is approximately 31.6 mL.