The analysis of some household cloudy ammonia cleaner involved diluting a 50.0mL sample of the cleaner to a volume of 250.0mL then titrating this diluted solution.
The diluted sample was found to have an NH3 (aq) concentration of 0.350 mole per L. What is the percentage by mass of NH3 (aq) in the original cloudy ammonia cleaner if it has a density of 0.986g per mL?
well, one diluted ti five times, so the original concentration must have been 5*.350M.
PercentMass= masssolute/masstotal * 100
= molesSolute*molmassNH3/(densitysolution*volumesolution) * 100
Need help to calculate chemistry problem
To solve this problem, we can follow the steps below:
Step 1: Calculate the original concentration of NH3 (aq) in the diluted solution.
Since the diluted solution was prepared by diluting the original cleaner by a factor of 5, the original concentration must have been 5 times the concentration of the diluted solution.
Original concentration = 5 * 0.350 M = 1.75 M
Step 2: Calculate the mass of NH3 (aq) in the diluted solution.
To calculate the mass, we need to know the molecular mass of NH3. The molecular mass of NH3 is 17.03 g/mol.
Mass of NH3 = concentration * volume * molecular mass
= 0.350 mol/L * 0.250 L * 17.03 g/mol
= 1.47575 g
Step 3: Calculate the mass of the original cloudy ammonia cleaner.
The mass of the original cleaner can be calculated using the density and volume information provided.
Mass of original cleaner = density * volume
= 0.986 g/mL * 50.0 mL
= 49.3 g
Step 4: Calculate the percentage by mass of NH3 (aq) in the original cleaner.
Percentage by mass = (mass of NH3 / mass of original cleaner) * 100
= (1.47575 g / 49.3 g) * 100
= 3.0 %
Therefore, the percentage by mass of NH3 (aq) in the original cloudy ammonia cleaner is approximately 3.0%.