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Most of the answers have been rounded up or rounded down to three significant figures (3sf)

Q21 ANSWERS

(a) mol HCl = 0.100 x 10.5/1000 = 0.00105

from the neutralisation equation Ca(OH)₂(aq) + 2HCl(aq) ===> CaCl₂(aq) + 2H₂O(l)

2 mol HCl needed to neutralise 1 mol Ca(OH)₂

therefore mol Ca(OH)₂ = 0.00105/2 = 0.000525 (in 25 cm³ or 25/1000 = 0.025 dm³)

Concentration of calcium hydroxide = 0.000525/0.025 = 0.0210 mol dm^-3

(b) (i) Mr[Ca(OH)₂] = 74, mass = mol x formula mass

mass Ca(OH)₂ in 1 dm³ = 0.021 x 74 = 1.554 g

Therefore concentration of Ca(OH)₂ = 1.55 g dm^-3

(ii) Since 100 cm3 is 1/10th of 1000 cm³ or 1 dm³

The concentration of Ca(OH)₂ = 0.155 g/100 cm^-3

Q22 ANSWER

NaOH + HCl ===> NaCl + H₂O

from equation mol NaOH = mol HCl = 0.100 x 20.55/1000 = 0.002055

25.0 cm³ = 25/1000 = 0.025 dm³

so, molarity NaOH = 0.002055/0.025 = 0.0822 mol dm^-3

Q23 ANSWER

mol NaOH in titration = 0.1025 x 17.65/1000 = 0.001809125

from the equation 2NaOH + H₂SO₄ ==> Na₂SO4 + 2H₂O

mol H₂SO₄ = mol NaOH/2

mol H₂SO₄ = 0.001809125/2 = 0.0009046 in 25.0/1000 = 0.0250 dm³

molarity H₂SO₄ = 0.0009046/0.025 = 0.0362 mol dm^-3 (3sf)

Q24 ANSWER

In the titration: mol NaOH = 0.1 x 22.5/1000 = 2.25 x 10^-3

Each mol of citric acid requires 3 mol of NaOH for complete neutralisation,

therefore, mol citric acid in 25.0 cm³ of the diluted solution = 2.25 x 10^-3/3 = 7.5 x 10^-4 mol

Since only 1/10th of the diluted solution was used in the titration ...

... the total moles of citric acid in the original cordial is 10 x 7.5 x 10^-4 = 7.5 x 10^-3 mol

M_r(citric acid, C₆H₈O₇) = 192, mass = mol x formula mass,

so mass of citric acid = 192 x 7.5 x 10^-3 = 1.44 g in 25.0/1000 = 0.0250 dm³

So the concentration of citric acid in the original cordial is 1.44/0.025 = 57.6 g/dm³

Q25 ANSWERS

Mass hydrated salt = 4.28, mass anhydrous salt = 1.89, mass water driven off = 4.28 - 1.89 = 2.39

M_r(H₂O) = 18, M_r(Na₂SO₄) = 142

mol H₂O = 2.39/18 = 0.1327, mol Na₂SO₄ = 1.89/142 = 0.01331

molar ratio H₂O/Na₂SO₄ = 9.97 ~10, therefore x = 10,

i.e. this hydrated form of sodium sulphate has the formula Na₂SO₄.10H₂O

Q26 ANSWER

the equation will be M + 2HCl ==> MCl₂ + H₂, from equation molar ratio M : H₂ of 1 : 1

therefore mol M = mol H₂ = 75/24000 = 3.125 x 10^-3

mol M = mass M/A_r(M), therefore A_r = mass M/mol M = 0.428/3.125 x 10^-3 = 136.96

A_r is ~137 which corresponds to barium Ba

Q27 ANSWER

Na₂CO₃ + 2HCl ==> 2NaCl + H₂O + CO₂

M_r(Na₂CO₃) = 106, so mol Na₂CO₃ titrated = 0.132/106 = 0.001245

from the molar equation, mol HCl = 2 x mol Na₂CO₃ = 0.00249

molarity HCl = mol/volume in dm3 = 0.00249/(24.8/1000) = 0.0996 mol dm^-3 (0.0996M)

Q28 ANSWERS

(a) Mr(Na₂CO₃) = 106, mol Na₂CO₃ = 1.30/106 = 0.01236 mol

volume = 250 cm³ = 25.0/1000 = 0.025 dm³

therefore molarity of Na₂CO₃ solution = 0.01236/0.025 = 0.04904 mol dm^-3 (4sf)

(b) mol Na₂CO₃ in each titration = 0.04904 x 25.0/1000 = 0.001226

from the molar equation Na₂CO₃ + 2HCl ==> 2NaCl + H₂O + CO₂

each mole of Na₂CO₃ needs two moles of HCl for complete neutralisation

therefore mol HCl = 0.001226 x 2 = 0.002452 in 24.35 cm³ (or 24.35/1000 = 0.02435 dm³)

therefore molarity HCl = 0.002452/0.02435 = 0.1007 mol dm^-3 (0.101 M 3sf)

 

Q29 ANSWERS

From equation mol Na₂CO₃ = mol HCl/2, M_r(Na₂CO₃) = 106, mass = mol x M_r

Analysis (i)

(a) mol HCl = 1.00 x 20.95/1000 = 0.02095, (b) mol Na₂CO₃ = 0.02095/2 = 0.010475

(c) mass titrated based on HCl titre = 0.010475 x 106 = 1.11035g, (d) % purity = 100 x 1.11035/1.113 = 99.76%

Analysis (ii)

(a) mol HCl = 1.00 x 20.55/1000 = 0.02055, (b) mol Na₂CO₃ = 0.02055/2 = 0.010275

(c) mass based on HCl titre = 0.010275 x 106 = 1.08915g, (d) % purity = 100 x 1.08915/1.092 = 99.74%

Analysis (iii)

(a) mol HCl = 1.00 x 21.90/1000 = 0.0219, (b) mol Na₂CO₃ = 0.0219/2 = 0.01095

(c) mass based on HCl titre = 0.01095 x 106 = 1.1607g, (d) % purity = 100 x 1.1607/1.166 = 99.54%

Most likely analysis result

All three values are reasonably close together, so the best estimate would be to average them

(99.76 + 99.73 + 99.55)/3 = 99.68%, 99.7% (1dp, 3sf) the analysis method is not really accurate enough for 4sf

In this method the main error is from the titration value, where an error of ≥0.05 cm³ is likely.

For a titration of 21 cm³, this equates to an error of 100 x 0.05/21 = 0.24%, so quoting beyond 3 significant figures or 1 decimal place is inappropriate i.e. the analytical result is best quoted as 99.7⁺/_-0.2%

Q30 ANSWER

Na₂CO₃ + 2HCl ==> 2NaCl + H₂O + CO₂, 1 mol hydrochloric acid relates to 0.5 moles of sodium carbonate

mol HCl used = 0.100 x 24.65/1000 = 0.002465, mol Na₂CO₃ titrated = 0.002465/2 = 0.0012325

M_r(Na₂CO₃) = 106, so mass Na₂CO₃`titrated = 106 x 0.0012325 = 0.130645

Therefore mass of H₂O in sample= 0.352 - 0.130645 = 0.221355 g

mole ratio Na₂CO₃ : H₂O is therefore 0.0012325 : 0.221355/18, giving 0.0012325 : 0.0122975

diving through by 0.0012325 gives a ratio of 1 : 9.98 (only a 0.2% error if x = 10)

Therefore the value of x can be reliably deduced as 10, since it would be expected to be an integer.

i.e. the formula of hydrated sodium carbonate crystals ('washing soda') is Na₂CO₃.10H₂O

Q31 ANSWER

(a) mass CuSO₄.xH₂O = 3.33, mass  anhydrous CuSO₄ = 2.13g

therefore mass of water driven off = 3.33 - 2.12 = 1.21 g

% water of crystallisation = 100 x 1.21/3.33 = 36.3% (63.7% is CuSO₄)

(b) M_r(H₂O) = 18,  M_r(CuSO₄) = 159.5

using the % composition from (a) you can calculate a mole ration based on 100g of the hydrated salt.

mol CuSO₄ = 63.7/159.5 = 0.399, mol H₂O = 36.3/18 = 2.016

the mole ratio H₂O/CuSO₄  = 2.016/0.399 = 5.05

To within a 1% error x = 5, so the formula of hydrated copper(II) sulfate is CuSO₄.5H₂O

(Note: assuming 5 is the answer, the error would be 100 x  (5.05 - 5.0)/5 = 1.0%, which would be quite acceptable.

Q32 ..

I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!