- Hydrobromic Acid (Molar Mass = 80.9 G/Mol) Is Commercially Available In A 34.0 Mass Percent Solution Which Has A Density Of 1.31 G/Cm². What Is The Molarity Of The Commercially Available Hydrobromic Acid?
The Correct Answer and Explanation is :
To determine the molarity of the commercially available hydrobromic acid (HBr), we use the formula: Molarity (M)=moles of solutevolume of solution in liters\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
Step 1: Calculate the mass of HBr in 1 liter of solution
The density of the solution is 1.31 g/cm³, equivalent to 1.31 g/mL.
Since 1 liter = 1000 mL: Mass of solution in 1 liter=1.31 g/mL×1000 mL=1310 g\text{Mass of solution in 1 liter} = 1.31 \, \text{g/mL} \times 1000 \, \text{mL} = 1310 \, \text{g}
The solution is 34.0% HBr by mass, meaning 34.0 g of HBr are present in 100 g of solution. Mass of HBr in 1 liter of solution=34.0100×1310=445.4 g\text{Mass of HBr in 1 liter of solution} = \frac{34.0}{100} \times 1310 = 445.4 \, \text{g}
Step 2: Convert the mass of HBr to moles
The molar mass of HBr is 80.9 g/mol. Moles of HBr=Mass of HBrMolar mass=445.480.9=5.51 moles\text{Moles of HBr} = \frac{\text{Mass of HBr}}{\text{Molar mass}} = \frac{445.4}{80.9} = 5.51 \, \text{moles}
Step 3: Calculate molarity
The volume of solution is 1 liter. Thus: Molarity (M)=Moles of HBrVolume of solution in liters=5.511=5.51 M\text{Molarity (M)} = \frac{\text{Moles of HBr}}{\text{Volume of solution in liters}} = \frac{5.51}{1} = 5.51 \, \text{M}
Final Answer:
The molarity of the commercially available hydrobromic acid is 5.51 M.
Explanation:
- Density and Percent Composition: The density allows us to calculate the mass of the solution per unit volume, while the percent composition gives the proportion of HBr in the solution.
- Moles of Solute: Using the molar mass, we convert the mass of HBr to moles.
- Molarity: Dividing the moles of HBr by the volume of the solution gives the molarity. This step links the amount of solute to its concentration in the solution.