An aqueas solution labeled 35% HClO4 had a density af 1.251g/Cm^3 calculate a) molarity b) molality
a) To calculate the molarity of the solution, we need to know the molecular weight of HClO4.
Molecular weight of HClO4 = 1(1) + 35.5 + 4(16) = 100.46 g/mol
Now, we can use the formula:
Molarity (M) = (mass percent / molecular weight) x (1,000 / density)
Substituting the values:
Molarity (M) = (35 / 100) x (1,000 / 100.46) x (1 / 1.251) = 9.87 M
Therefore, the molarity of the solution is 9.87 M.
b) To calculate the molality of the solution, we need to convert the molarity into moles of solute per kilogram of solvent.
First, we need to calculate the mass of the solution:
Mass of solution = volume x density
Assuming a volume of 1 L, then
Mass of solution = 1 x 1.251 = 1.251 kg
Now, we can calculate the moles of HClO4 in 1 L of the solution:
Moles of HClO4 = 9.87 x 1 = 9.87 mol
As the density and molecular weight of water are very close to 1 g/cm³ and 18 g/mol, respectively, we can assume that 1 L of water weighs 1 kg.
Therefore, the molality of the solution is:
Molality = moles of HClO4 / mass of water in kg
Molality = 9.87 / 1 = 9.87 mol/kg
Therefore, the molality of the solution is 9.87 mol/kg.
To calculate the molarity and molality of the 35% HClO4 solution, we will need to know the molecular weight (molar mass) of HClO4. The molecular weight of HClO4 is 100.46 g/mol.
a) To calculate the molarity (M), we need to determine the number of moles of HClO4 present in the solution.
First, we need to determine the mass of HClO4 in the solution. The solution is labeled as 35% HClO4, which means that 35 g of HClO4 is present in 100 g of the solution.
mass of HClO4 = 35% * 100 g = 35 g
We also know the density of the solution, which is 1.251 g/cm^3. This density value can be used to convert the mass of the solution into volume using the relationship:
density = mass/volume
Rearranging this equation, we have:
volume = mass/density
volume = 100 g / 1.251 g/cm^3 = 79.936 cm^3 = 0.079936 L (since 1 cm^3 = 0.001 L)
Next, we can calculate the number of moles of HClO4 using the mass and molecular weight:
moles of HClO4 = mass/molecular weight
= 35 g / 100.46 g/mol
= 0.3485 mol
Finally, we can calculate the molarity by dividing the moles of HClO4 by the volume of the solution in liters:
molarity (M) = moles/volume
= 0.3485 mol / 0.079936 L
= 4.359 M
Therefore, the molarity of the 35% HClO4 solution is approximately 4.359 M.
b) To calculate the molality (m), we need to determine the number of moles of HClO4 present per kilogram of solvent (water).
Since we know the mass of HClO4 is 35 g, and the density of the solution is 1.251 g/cm^3, we can calculate the mass of the solution (water + solute):
mass of solution = mass of water + mass of HClO4
mass of water = mass of solution - mass of HClO4
= 100 g - 35 g
= 65 g
Next, we convert the mass of water to kilograms:
mass of water = 65 g * (1 kg / 1000 g)
= 0.065 kg
Finally, we can calculate the molality by dividing the moles of HClO4 by the mass of water in kilograms:
molality (m) = moles of solute / mass of solvent (in kg)
= 0.3485 mol / 0.065 kg
= 5.363 mol/kg
Therefore, the molality of the 35% HClO4 solution is approximately 5.363 mol/kg.