Question

1. Calculate the molarity of a solution made by adding 20.8 mL of concentrated potassium hydroxide (34.5 % by mass, density 1.46 g/mL) to some water in a volumetric flask, then adding water to the mark to make exactly 500 mL of solution. (It is important to add concentrated acid or base to water, rather than the other way, to minimize splashing and maximize safety.)

[KOH]= _________ M

2. A 0.450-gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO₃(s) + 2HCl(aq) ---> CaCl₂(aq) + H₂O(l) + CO₂(g)

The excess HCl(aq) is titrated by 9.10 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample.

__________ %

Answer 1

Calculation of the molarity of the stock solution of KOH :-

(1)Calculation of moles of KOH in 100.0 g of 34.5% solution :

mass of KOH = 34.5 g

molarmass of KOH = 39 + 16 + 1 = 56 g/mol

So number of moles of KOH is = mass/molar mass

                                             = 34.5 g / 56 (g/mol)

                                            = 0.616 mol

(2) calculation of volume of 100.0 g of solution :

density = mass / volume

Volume of the solution is , Volume = mass / density

                                                  = 100 g / 1.46 (g/mL)

                                                 = 68.5 mL

                                                = 0.0685 L

(3) Calculation of molarity :-

Molarity , M = number of moles / volume in L

                  = 0.616 mol / 0.0685 L

                  = 9.0 M

So the molarity of stock solution is 9.0 M

According to law of dilution MV = M'V'

Where

M= molarity of stock solution = 9.0 M

M' = molarity of dilute solution = ?

V= volume of stock solution taken = 20.8 mL

V' = Volume of dilute solution = 500 mL

Plug the values we get

So the molarity of dilute solution is 0.374 M

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Number of moles of HCl is , n = molarity x volume in L

                                           = 0.150 M x 50.0x10^-3 L

                                          = 0.0075 moles

Number of moles of NaOH is , n' = Molarity x volume in L

                                               = 0.125 M x 9.10 x10-3 L

                                              = 0.001137 moles

HCl + NaOH NaCl + H₂O

So 1 mole of NaOH reacts with 1 mole of HCl

    0.001137 moles of NaOH reacts with 0.001137 moles of HCl

So 0.0075 - 0.001137 moles = 0.00636 moles of HCl reacted with the sample containing CaCO₃

CaCO₃(s) + 2HCl(aq) CaCl₂(aq) + H₂O(l) + CO₂(g)

According to the balanced Equation ,

2 moles of HCl reacts with 1 mole of CaCO₃

0.00636 moles of HCl reacts with (0.00636 / 2) moles of CaCO₃

                                                 = 0.00318 moles of CaCO₃

Molar mass of CaCO₃ is = At.mass of Ca + At.mass of C + (3xAt.mass of O)

                                    = 40 + 12 + (3x16)

                                    = 100 g /mol

So mass of CaCO₃ is ,m = number of moles x molar mass

                                     = 0.00318 mol x 100 g/mol

                                     = 0.318 g

So percentage of CaCO₃ in the sample = ( mass of CaCO₃ / mass of the impure sample ) x 100

                                                         = ( 0.318 / 0.450 ) x100

                                                         = 70.7 %