Molarity Calculation for Strontium Chloride Solution

Molarity Problem Sample

Problem Statement

  • 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.

Definitions

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

Steps to Solve the Problem

  1. **Calculate Moles of Strontium Chloride:

    • Use the formula to find the number of moles (n):**
      n=massmolar massn = \frac{mass}{molar \ mass}
    • For this problem:
      n=5 g158.53 g/moln = \frac{5\text{ g}}{158.53\text{ g/mol}}
    • Calculate the moles:
      n≈0.0316 molesn \approx 0.0316 \text{ moles}

  2. Determine Volume of Solution in Liters needed to create a 1 M Solution:

    • Rearrange the molarity equation to find the volume (V):
      V=nMV = \frac{n}{M}
    • Substitute the values:
      V=0.0316 moles1 MV = \frac{0.0316 \text{ moles}}{1\text{ M}}
    • Calculate the volume:
      V≈0.0316 LV \approx 0.0316 \text{ L}

  3. Convert Volume to Milliliters:

    • To convert liters to milliliters, multiply by 1000.
    • (0.0316extL)(1000extmL/L)≈31.6extmL(0.0316 ext{ L}) (1000 ext{ mL/L}) \approx 31.6 ext{ mL}

Final Solution

  • 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.