An aqueas solution labeled 35% HClO4 had a density af 1.251g/Cm^3 calculate a) molarity b) molality

Answer 1

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.