Find the molarity of 37 wt% HCl that has a density is 1.19 g/mL. How much volume is needed to prepare 100 mL of 6 M of HCl from the concentrated solution?
The Correct Answer and Explanation is:
To solve this problem, we need to first find the molarity of the concentrated HCl solution and then calculate how much volume of that solution is required to prepare 100 mL of a 6 M HCl solution.
Step 1: Find the molarity of the 37 wt% HCl solution.
We are given:
- Weight percentage of HCl = 37%
- Density of the solution = 1.19 g/mL
- Calculate the mass of 1 liter of solution:
Since the density is 1.19 g/mL, 1 liter (1000 mL) of the solution weighs: mass of solution=1.19 g/mL×1000 mL=1190 g\text{mass of solution} = 1.19 \, \text{g/mL} \times 1000 \, \text{mL} = 1190 \, \text{g}mass of solution=1.19g/mL×1000mL=1190g - Calculate the mass of HCl in 1 liter of solution:
37% of the mass is HCl, so: mass of HCl=0.37×1190 g=440.3 g\text{mass of HCl} = 0.37 \times 1190 \, \text{g} = 440.3 \, \text{g}mass of HCl=0.37×1190g=440.3g - Calculate the moles of HCl:
The molar mass of HCl is approximately 36.46 g/mol. So: moles of HCl=440.3 g36.46 g/mol≈12.08 mol\text{moles of HCl} = \frac{440.3 \, \text{g}}{36.46 \, \text{g/mol}} \approx 12.08 \, \text{mol}moles of HCl=36.46g/mol440.3g≈12.08mol - Calculate the molarity of the concentrated HCl solution:
The molarity MMM is the number of moles of solute per liter of solution: M=moles of HClvolume of solution in liters=12.08 mol1 L=12.08 MM = \frac{\text{moles of HCl}}{\text{volume of solution in liters}} = \frac{12.08 \, \text{mol}}{1 \, \text{L}} = 12.08 \, \text{M}M=volume of solution in litersmoles of HCl=1L12.08mol=12.08M So, the molarity of the concentrated solution is approximately 12.08 M.
Step 2: Calculate the volume needed to prepare 100 mL of 6 M HCl solution.
We can use the dilution equation:M1V1=M2V2M_1 V_1 = M_2 V_2M1V1=M2V2
where:
- M1M_1M1 = 12.08 M (the concentration of the stock solution)
- V1V_1V1 = the volume of the stock solution needed
- M2M_2M2 = 6 M (the concentration of the final solution)
- V2V_2V2 = 100 mL (the volume of the final solution)
Rearranging the dilution equation to solve for V1V_1V1:V1=M2V2M1=6 M×100 mL12.08 M≈49.7 mLV_1 = \frac{M_2 V_2}{M_1} = \frac{6 \, \text{M} \times 100 \, \text{mL}}{12.08 \, \text{M}} \approx 49.7 \, \text{mL}V1=M1M2V2=12.08M6M×100mL≈49.7mL
Final Answer:
- The molarity of the 37 wt% HCl solution is approximately 12.08 M.
- To prepare 100 mL of a 6 M HCl solution, you would need approximately 49.7 mL of the concentrated HCl solution.
This process involves using the density to find the mass of the concentrated solution, determining the moles of HCl, and then applying the dilution equation to find the necessary volume.
