Question

A 5.00 mL sample of toilet cleaner weighs 4.902 g. When titrated, those 5.00 mL of the household product need 25.50 mL of NaOH 0.506 M for the end point to be reached.

a)    What is the density of the toilet cleaner?

b)    How many moles of HCl are contained in those 5.00 mL of household product?

c)    What mass of HCl is contained in those 5.00 mL of household product?

d)    What is the molarity of the toilet cleaner?

e)    What is the mass percentage of HCl in the toilet cleaner?

Answer 1

a) density of the toilet cleaner = mass/volume

     = 4.902/5.00 = 0.98 g/ml

b) no of moles of HCl are contained in those 5.00 mL of household product = no of mol of NaOH reacted

   = 25.5*0.506/1000

   = 0.0129 mol

c) mass of HCl = n*mWT

   = 0.0129*36.5

   = 0.471 g

d) molarity = (w/mwt)*(1000/v in ml)

         = (0.471/36.5)*(1000/5)

   = 2.58 M

e) mass% of cleaner = M*Mwt/d*10

             = 2.58*36.5/(0.98*10)

             = 9.61%