The density of the concentrated HBr solution of 48% by mass is 1.50 g / ml. a) How many grams of concentrated HBr should be used to prepare 250 ml of 4.0M HBr solution? b) How many ml of concentrated HBr should be used to prepare this solution. Br: 80 g / mol H: 1 g / mol
The density of the concentrated HBr solution of 48% by mass is 1.50 g / ml.ÂÂ
a) How many grams of concentrated HBr should be used to prepare 250 ml of 4.0M HBr solution?
b) How many ml of concentrated HBr should be used to prepare this solution.
The Correct Answer and Explanation is :
Given Data:
- Concentration of HBr solution = 48% by mass.
- Density of concentrated HBr = 1.50 g/mL.
- Desired solution: 250 mL of 4.0 M HBr.
- Molar mass of HBr = 80+1=81 g/mol80 + 1 = 81 \, \text{g/mol}.
Step 1: Calculate the moles of HBr needed
Using the molarity equation: Molarity (M)=moles of solutevolume of solution in liters\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
Rearranging to find moles: Moles of HBr=4.0 M×0.250 L=1.0 mol.\text{Moles of HBr} = 4.0 \, \text{M} \times 0.250 \, \text{L} = 1.0 \, \text{mol}.
Step 2: Calculate the mass of HBr required
Using the molar mass of HBr (81 g/mol81 \, \text{g/mol}): Mass of HBr required=1.0 mol×81 g/mol=81 g.\text{Mass of HBr required} = 1.0 \, \text{mol} \times 81 \, \text{g/mol} = 81 \, \text{g}.
Step 3: Calculate the mass of concentrated HBr solution needed
The concentrated solution is 48% HBr by mass, meaning 48 g of HBr is present in every 100 g of solution. Using this proportion: Mass of concentrated HBr solution=Mass of HBr requiredMass fraction of HBr=810.48=168.75 g.\text{Mass of concentrated HBr solution} = \frac{\text{Mass of HBr required}}{\text{Mass fraction of HBr}} = \frac{81}{0.48} = 168.75 \, \text{g}.
Step 4: Calculate the volume of concentrated HBr solution needed
The density of the concentrated HBr solution is 1.50 g/mL1.50 \, \text{g/mL}. Using the density formula: Volume of concentrated solution=Mass of concentrated solutionDensity=168.751.50=112.5 mL.\text{Volume of concentrated solution} = \frac{\text{Mass of concentrated solution}}{\text{Density}} = \frac{168.75}{1.50} = 112.5 \, \text{mL}.
Final Answers:
a) 168.75 g of concentrated HBr solution is required.
b) 112.5 mL of concentrated HBr solution is required.
Explanation (300 Words):
The problem requires preparing 250 mL of 4.0 M HBr solution using a 48% concentrated HBr solution. The solution’s molarity (M) determines how many moles of solute are needed in a specific volume. To start, we calculated the moles of HBr required using the formula Moles=Molarity×Volume (in liters)\text{Moles} = \text{Molarity} \times \text{Volume (in liters)}, giving 1.0 mol1.0 \, \text{mol} of HBr.
Next, we converted moles to grams using HBr’s molar mass (81 g/mol81 \, \text{g/mol}), obtaining 81 g81 \, \text{g} of pure HBr. However, the concentrated HBr solution is only 48% HBr by mass, meaning it contains 48 g48 \, \text{g} of HBr per 100 g100 \, \text{g} of solution. To find the total mass of concentrated solution required to provide 81 g81 \, \text{g} of HBr, we divided 81 g81 \, \text{g} by the mass fraction 0.480.48, resulting in 168.75 g168.75 \, \text{g}.
Finally, since the density of the concentrated solution is 1.50 g/mL1.50 \, \text{g/mL}, we converted the mass of the concentrated solution into volume using Volume=MassDensity\text{Volume} = \frac{\text{Mass}}{\text{Density}}. This calculation gave 112.5 mL112.5 \, \text{mL}, the volume of concentrated HBr solution needed.
These steps ensure precision in preparing the desired solution, adhering to the principles of solution chemistry.