Commercial hydrobromic acid can be purchased as 40.0 % by mass 9f HBr. The density of this acid is 1.38 g m/L. Calculate the molality of this commercial acid..please show me the solution..thanks
As in the HCl problem, start with the definition.
m = mols/kg solvent
mass of 1 L = 1000 mL x 1.38 g/mL = 1380 g.
Not all of that is HBr. HBr present is 40% so 1380 x 0.40 = 552g.
mols in 552g HBr is mols = grams/molar mass = 552/about 81 (but you need to do it more accurately).
mols HBr = estd 6.81.
How much solvent is there? You have 1380g solution of which 552g is HBr; therefore, g solvent 1380-552 = 828g.
Convert 828g to kg, then m = mols/kg solvent = ?m
Sure, here's the solution with a touch of humor:
To calculate the molality of this commercial acid, we need to first convert the given mass percentage of HBr to grams of HBr.
Assuming we have 100 grams of the commercial acid, that means we have 40 grams of HBr (40.0% x 100g).
Now, we need to find the moles of HBr. We can do this by dividing the mass of HBr by its molar mass. The molar mass of HBr is 1 (atomic mass of H) + 79.9 (atomic mass of Br) = 80.9 g/mol.
So, the number of moles of HBr is 40 g / 80.9 g/mol ≈ 0.494 moles.
Now, we need to determine the mass of the solvent (water) to calculate the molality. Here's where density comes into play.
Given that the density of the acid is 1.38 g/mL, and assuming we have 1000 mL (1 L) of the acid (since the density is given in g/mL), we can multiply the density by the volume to find the mass. Thus, the mass of the acid is 1.38 g/mL x 1000 mL = 1380 g.
Subtracting the mass of HBr (40 g), we find that the mass of the solvent (water) is 1380 g - 40 g = 1340 g.
Finally, we can calculate the molality by dividing the moles of HBr by the mass of the solvent in kilograms.
Molality = moles of solute / mass of solvent (in kg)
Converting the mass of the solvent to kg gives us 1340 g / 1000 g/kg = 1.34 kg.
Thus, the molality of the commercial acid is 0.494 moles / 1.34 kg ≈ 0.369 mol/kg.
So, there you have it! The molality of the commercial hydrobromic acid is approximately 0.369 mol/kg. Keep in mind that this calculation assumes you have 100 grams of the commercial acid. Be sure to adjust accordingly based on the actual amount you have.
To calculate the molality of the commercial hydrobromic acid, you need to first find the molality formula, which is defined as the moles of solute per kilogram of solvent.
Step 1: Find the mass of HBr in the hydrobromic acid solution.
Since the hydrobromic acid is 40.0% by mass HBr, for every 100 g of the solution, 40.0 g is HBr.
Therefore, the mass of HBr in the solution is (40.0 g).
Step 2: Convert the mass of HBr to moles.
To convert the mass of HBr to moles, we need to use the molar mass of HBr, which is 80.91 g/mol. The molar mass is obtained by adding the atomic masses of hydrogen (1 g/mol) and bromine (79.91 g/mol).
To find the number of moles, divide the mass of HBr by its molar mass:
Number of moles = (mass of HBr) / (molar mass of HBr) = (40.0 g) / (80.91 g/mol) ≈ 0.495 mol
Step 3: Calculate the mass of the solvent (water).
To calculate the mass of the solvent (water), we need to use the density of the acid, which is given as 1.38 g/mL. Since 1 mL is equal to 1 gram of water, the density can be expressed as 1.38 g/mL = 1.38 g/cm³.
Since the density of water is approximately 1 g/mL or 1 g/cm³, the volume of water is equal to its mass in grams. Thus, the mass of the solvent (water) is 1000 g (1 L).
Step 4: Calculate the molality.
The molality (m) is defined as moles of solute per kilogram of solvent, so we need to convert the mass of the solvent (water) to kilograms.
The mass of the solvent is 1000 g, which is equal to 1 kg.
Divide the number of moles of HBr by the mass of the solvent in kilograms to get the molality:
Molality (m) = (moles of solute) / (mass of solvent in kg) = 0.495 mol / 1 kg ≈ 0.495 mol/kg
Therefore, the molality of the commercial hydrobromic acid is approximately 0.495 mol/kg.
To calculate the molality of the commercial hydrobromic acid, you need to know the formula weight of HBr, the mass percent of HBr in the solution, and the density of the solution.
First, find the formula weight of HBr using the atomic masses of hydrogen and bromine:
HBr = (1.00784 g/mol) + (79.904 g/mol) = 80.91184 g/mol
Next, convert the mass percent of HBr to grams by assuming you have 100 g of the solution:
Mass of HBr = (40.0 g HBr/100 g solution) × 100 g solution = 40.0 g HBr
Now, we need to find the moles of HBr:
Moles of HBr = Mass of HBr / Formula weight of HBr
Moles of HBr = 40.0 g / 80.91184 g/mol = 0.493 mol HBr
Finally, calculate the molality of the commercial acid using the following formula:
Molality = Moles of solute / Mass of solvent (in kg)
Since the density is given in g/mL, we can convert it to g/L by multiplying by 1000:
Density = 1.38 g/mL × 1000 mL/L = 1380 g/L
Then, divide the mass of HBr by the density of the solution to get the mass of the solvent:
Mass of solvent = Mass of solution - Mass of HBr
Mass of solvent = 100 g - 40.0 g = 60.0 g
Converting the mass of the solvent to kg:
Mass of solvent = 60.0 g × (1 kg / 1000 g) = 0.06 kg
Finally, plug the values into the formula for molality:
Molality = 0.493 mol HBr / 0.06 kg = 8.22 mol/kg.
Therefore, the molality of the commercial hydrobromic acid is 8.22 mol/kg.