WAEC SSCE - Chemistry - 2024 (Practical)

Detalhes da Resposta

(a) Volume of pipette used = 25.00 cm\(^3\)

Volume of A used =  \(\frac{Rough + 1st titre + 2nd titre}{3}\)

                             = \(\frac{20.90 + 20.80 + 21.00}{3}\)

                             = 20.90 cm\(^3\)

(b)  2HNO\(_3\)\(_{aq}\)  +  Na\(_2\)CO\(_3\).xH\(_2\)O\(_{aq}\) → 2NaNO\(_3\) + CO\(_2\) + (X+1)H\(_2\)0(l)

c(i) concentration of E in moldm\(^{-3}\)

Since we do not know the concentration of D in moldm\(^{-3}\), but in g per cm\(^3\), it will be difficult using the fomula \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\).

Thus, we must first convert the Conc. of D to moldm\(^{-3}\).

We were told from the question that  D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\),

So, if D            6.30 g of HNO\(_3\) was dissolved in 500 cm\(_3\)

then,                 X g of HNO\(_3\) will be dissolved in 1000 cm\(_3\)

   X =  \(\frac{{6.30}\times{1000}}{500}\) = 12.6 gdm\(^{-3}\)

Recall that  Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)

Molar mass od HNO\(_3\) = 1 + 14 + (16 X 3) = 63 g/mol

concentration of D(moldm\(^{-3}\)) =  \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) 

                                                  C\(_D\)     =  \(\frac{12.6}{63}\) = 0.20 moldm\(^{-3}\)

Now we can use the formula \(\frac{C_D V_D}{C_E V_E}\) = \(\frac{N_D}{N_E}\).

Given that 

C\(_D\) =   0.20 moldm\(^{-3}\)                                            N\(_D\) = 2

V\(_D\) =  20.90cm\(^3\)                                                      N\(_E\) = 1

C\(_E\) = ?

V\(_E\) =  25.0cm\(^3\)

 C\(_B\) = \(\frac{C_D V_D N_E}{N_D V_E}\)

                     C\(_B\) = \(\frac{ 0.2 X 20.90 X 1}{25 X 2}\)

                                 = 0.0836 moldm\(^{-3}\) 

                                 = 0.084 moldm\(^{-3}\) 

(ii) To get the value of x in Na\(_2\)CO\(_3\).xH\(_2\)O

First, from the question, if   8.20 g of Na\(_2\)CO\(_3\).xH\(_2\)O was dissolved in 250 cm\(^3\) of distilled water.

 then,                                    Xg of Na\(_2\)CO\(_3\).xH\(_2\)O will be dissolved in 1000 cm\(^3\) of distilled water.

So,  X =  \(\frac{{8.20}\times{1000}}{250}\) = 32.8 gdm\(^{-3}\)

Mass concentration of E = 32.8 gdm\(^{-3}\)

Recall that  concentration of E(moldm\(^{-3}\)) =  \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) 

Then,  Molar mass of E ( Na\(_2\)CO\(_3\).xH\(_2\)O ) =  \(\frac{Mass concentration}{Molar concentration}\) 

                                                                                       =  \(\frac{32.8}{0.0836}\) 

                                                                                       = 392.34 g/mol

But    Molar mass of Na\(_2\)CO\(_3\).xH\(_2\)O = Na\(_2\)CO\(_3\) + x(H\(_2\)O)

                                                               392.34   = (23 x 2) + 12 + (16 x 3) + 18x

                                                               392.34  =  46 + 12 + 48 + 18x

                                                               392.34 =  106 + 18x

                                                                 18x = 392.34 - 106

                                                                  18x = 286.34

                                                                   x = \(\frac{286.34}{18}\)

                                                                   x = 15.9

                                                                   x ≈ 16

Detalhes da Resposta

(a) Volume of pipette used (Volume of B used) = 25.0 cm\(^3\)

Volume of A used =  \(\frac{Rough + 1st titre + 2nd titre}{3}\)

                             = \(\frac{18.20 + 18.10 + 18.20}{3}\)

                             = 18.166

                 V\(_A\) = 18.17 cm\(^3\)

(i) Concentration of B in moldm\(^{-3}\) 

  From the balanced equation :   MnO\(_4\)\(^-\)\(_{(aq)}\) +  5Fe\(^{2+}\)\(_{(aq)}\)  +  8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\)  + Fe\(^{3+}\)\(_{(aq)}\) +  4H\(_2\)O\(_{(l)}\)   

C\(_A\) =   0.020 moldm\(^{-3}\)                                            N\(_A\) = 1

V\(_A\) =  18.17cm\(^3\)                                                        N\(_B\) = 5

C\(_B\) = ?

V\(_B\) =  25.0cm\(^3\)

Using   \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\)

                     C\(_B\) = \(\frac{C_A V_A N_B}{N_A V_B}\)

                     C\(_B\) = \(\frac{ 0.02 X 18.17 X 5}{25 X 1}\)

                     C\(_B\) = 0.07268

                     C\(_B\) = 0.070 moldm\(^{-3}\)

(ii) Concentration of B in gdm\(^{-3}\)

     If  3.8 g of Fe granules was dissolved in 250 cm\(^3\) of B

          Xg of Fe granules will be dissolved in 1000 cm\(^3\) of B

          X = \(\frac{ 3.8 X 1000}{250}\) =  15.2 gdm\(^{-3}\)

    Concentration of B in gdm\(^{-3}\) will be 15.2 gdm\(^{-3}\)

(iii) Mass of Fe\(^{2+}\) in 250cm\(^3\) of B.

 To get the mass, we'll use the formula  n = \(\frac{mass}{Molar mass}\)

However, since do not know the number of moles, n , hence we'll need to obtain n, using  n =  \(\frac{CV}{1000}\) since we have C and V

So,  n =  \(\frac{{0.070}\times{250}}{1000}\) =  0.025 mole

Now, we can apply n = \(\frac{mass}{Molar mass}\) to obtain the mass, given that the Molar mass of Fe was given to be 56

                               n = \(\frac{mass}{Molar mass}\)

                               m = n x M = 0.018 x 56 = 1.02 g

Therefore, 1.02 g of Fe\(^{2+}\) in 250 cm\(^3\) of B.

(iv) Percentage of Fe\(^{2+}\) in the granules =  \(\frac{Mass of Fe^{2+}}{Total mass of Fe granules}\times 100\)

                                                                       = \(\frac{1.02}{3.8}\times 100\)

                                                                       = 26.89%

                                                                       = 26.9%

Detalhes da Resposta

Detalhes da Resposta

(a) Gases that are soluble in water are:

- Ammonia

- Hydrogen chloride 

- oxygen,

- carbon dioxide,

- Sulphur dioxide,

- chlorine, and

- hydrogen sulfide. 

(b) Apparatus that could be used to measure volume of solutions accurately:

- volumetric flask,

- graduated cylinder,

- burette, or

- volumetric pipette. 

(c) In a laboratory, fume cupboards are essential safety equipment used to capture and remove airborne hazardous or poisonous substances like gases, vapours, aerosols, and dust, protecting users from exposure to harmful chemicals. 

(d) Personal protective equipment (PPE) used in the laboratory.

- Lab coats

- safety glasses

- respiratory protector

- Hand gloves

- Face shield

Detalhes da Resposta

a(i) Salt solution is acidic

     - Reaction is exothermic.

    - White precipitate is likely to be a trioxocarbonate(IV) salt of Al, Ca and Zn.

    - Prepared solution contained soluble salt of Al, Ca and Zn.       

b(i)PbSO\(_4\)

  (ii) NaHCO\(_3\)

  (iii) NaHCO\(_3\) ; Cu(NO\(_3\))\(_2\).

  (iv) NaHCO\(_3\)

  (v) NH\(_4\)Cl

(c) Precautions to be taken during titration include:

- Wear a lab coat to protect your clothing from spills. 

- Ensure to rinse the burette with an acid solution

- Also ensure the funnel is removed immediately after filling the acid into the burette.

- Avoid error due to parallax while taking the reading to determine the end point.

- Ensure all glassware is clean and free of bubbles. 

Detalhes da Resposta

F = (NH\(_4\))\(_2\)CO\(_3\)