Question

3. How many moles of bleach are in each trial if you use 5.00 mL of a 6.00% (m/m) solution? Use the density you found in Question 2.
this is the question

this is thr data

please help thank you

Answer 1

Ans:

As per the question, 5 ml of 6% (m/m) bleach solution is used

Trial 1

For Trial 1, d = 1.12 g / ml

Therefore, mass = Volume x Density = 5 x 1.12 = 5.6 g of solution
mass of solution = 5.6g
% of solute in solution is 6% (m/m)
Therefore mass of solute (bleach) = 6% of 5.6g = 0.336g of solute.

Moles of Bleach present = (0.336g) / 74.44 = 0.00451 moles (Molecular weight of bleach = 74.44 g/mol)

Thus 0.00451 moles of bleach is present in this trial.

Trial 2

For Trial 2, d = 1.125 g / ml

Therefore, mass = Volume x Density = 5 x 1.125 = 5.625 g of solution
mass of solution = 5.625g
% of solute in solution is 6% (m/m)
Therefore mass of solute (bleach) = 6% of 5.625g = 0.337g of solute.

Moles of Bleach present = (0.337g) / 74.44 = 0.00453 moles (Molecular weight of Bleach = 74.44 g/mol)

Thus 0.00453 moles of bleach is present in this trial.

Trial 3

For Trial 3, d = 1.0925 g / ml

Therefore, mass = Volume x Density = 5 x 1.0925 = 5.462 g of solution
mass of solution = 5.462g
% of solute in solution is 6% (m/m)
Therefore mass of solute (bleach) = 6% of 5.462g = 0.327g of solute.

Moles of Bleach present = (0.327g) / 74.44 = 0.0044 moles (Molecular weight of Bleach = 74.44 g/mol)

Thus 0.00440 moles of bleach is present in this trial.

Trial 4

For Trial 4, d = 1.105 g / ml

Therefore, mass = Volume x Density = 5 x 1.105 = 5.525 g of solution
mass of solution = 5.525g
% of solute in solution is 6% (m/m)
Therefore mass of solute (bleach) = 6% of 5.525g = 0.331g of solute.

Moles of Bleach present = (0.3315g) / 74.44 = 0.0044 moles (Molecular weight of Bleach = 74.44 g/mol)

Thus 0.00445 moles of bleach is present in this trial.