Quantitative Chemistry: Stoichiometry and Solutions

The resulting solution has still one mole H⁺ ion to be neutralised, so, 1 mole of NaOH is required for complete neutralisation. Hence, this is a correct option.

The required reaction is, MnO₄^– +  8H⁺ +5e  →  Mn²⁺ + 4H₂O
Cu⁺ – e → Cu²⁺
S^2– – 6e → S⁴⁺ Therefore, eq. wt. of Cu₂S = (M1 / ((1*2) + 6)) = M1 / 8
Hence, this is a correct option.
 

Let there be x ml CO in the mixture. Hence, there will be (1000 – x) ml CO₂ . The reaction of CO₂ with red hot charcoal may be given as,CO₂ (g) + C(s) → 2CO(g)from reaction we have, x = 400 ml
Therefore volume of CO = 400 ml and hence the volume of CO₂ = (1000 – 400 ) ml = 600 ml
Hence, this is the correct option.

We know that,
No. of g-equivalents = normality * volume (L)
Therefore, no. of g-equivalent of HCl = ((5 * 100) / 1000) = 0.5, which is the required answer; hence this is a correct option.

No. of moles CO2 = (4.4 / 44) = 0.1
No. of molecules = 0.1 * 6.022 * 10^23
= 6.022 * 10^22, which match with the given option, hence this is a correct option.

Milli eq. of HCl initially = 10 * 0.5 = 5.
Milli eq. of NaOH consumed = milli eq. of HCl in excess     = 10 * 0.2 = 2
Therefore, milli eq. of HCl consumed = Millie q. of Ca(OH)₂     = 5 – 2 =3
Therefore, eq. of Ca(OH)₂ = (3 / 1000) = 3 * 10^–3
Mass of Ca(OH)₂ = (3 * 10^–3) * (74 / 2) = 0.111 g% Ba(OH)₂ = (0.111 / 10) * 100 = 1.11%; which is the required solution of the given problem. Hence, this is the correct option.

Mass of 1 litre of H₂SO₄ solution = volume * density
= 1000 *1.84 g = 1840 g
= mass of H₂SO₄ present in 1 litre 96% H₂SO₄ solution
= (96 / 100) *1840 g = 1766.4 g
Or, strength of H₂SO₄ solution = 1766.4 g / litre
Or Normality (N) = (Strength (gL^-1) / equivalent Mass)  
= (1766.4 / 49) = 36.05 (N); which match with the given option, hence this is a correct option.

The equiv. of H⁺ in 30 ml of (N / 2) HCl = (10 / 2) 10^–3
The equiv. of H⁺ in 30 ml of (N / 10) HNO3 = (30 / 10) * 10^–3
The equiv. of H⁺ in 75 ml of (N / 10) HNO3 = (75 / 5) * 10^–3.
Hence, total equiv. of H⁺ (5 + 3 + 15) * 10^–3 = 23 10^–3
Total volume of solution = 115 ml
Hence, normality of H⁺ in the resulting mixture = [(23 * 10^–3 * 10^3) / 115) (N) = (N / 5) = 0.2 (N).
Hence, this is a correct option.

The required reaction is S + O₂→ SO₂ + Cl₂ (water) → SO₄^2– + BaCl₂→BaSO₄ +2Cl^–. Moles of sulphur = (8 / 32) = (1 / 4),  moles of SO2. Therefore, moles of BaSO₄ precipitated = 0.25. (applying principle of atom conservation). Hence, this is a correct option.

Mol. mass of CaCO₃ = 100
Mol. mass of Ca = 40
20 mg of CaCO₃ has ((4020) / 100) = 8 mg of Ca⁺² ions
Since, 8 mg of Ca⁺² in 100 L sample of hard water. 101000 ml sample of hard water has 8 mg of Ca⁺² therefore, 10^6 ml sample of hard water has 80 mg of Ca⁺²,i.e., 80 ppm, which is the required solution of the given problem. Hence, this is a correct option.

Molecular mass of NaBrO₃ = 151
Each bromate ion takes up 6 electrons; therefore
Eq. mass of NaBrO₃ = Mol. mass / 6 = 151 / 6
Amount of NaBrO₃ in 85.5 ml 0.672 N solution= (0.672 / 1000) * (151 / 6) * (85.5) g = 1.446 g.
Molarity = (Normality / nf) = 0.672 / 6 = 0.112 M.
Hence, this is a correct option.

The reaction can be given as: Reduction half→Cr₂O₇^2– + 14H⁺ + 6e →  2Cr³⁺ + 7H₂O
Oxidation half→[Sn²⁺ – 2e → Sn⁴⁺]*3
So the balanced chemical equation is
3Sn²⁺ + Cr₂O₇^2– +14H⁺ = 3Sn⁴⁺ + 2Cr³⁺ + 7H₂O
therefore, x + y + z = 18, hence this is a correct option.
 

Molecular mass of the mixture = 2 * 38.3 = 76.6
Let, mass of NO₂ in 100 g of mixture = a g.
Moles of NO₂ + Moles of N₂O₄ = Moles of mixture
i.e., (a / 46) + ((100 – a ) / 92) = (100 / 76.6)
a = 30.1 g.
Moles of NO₂ in the mixture =( 20.1 / 46) = 0.437 (approx.).
Hence, this is a correct option.

Mass of sugar = 34.2 g
No. of moles of sugar = (34.2 / Mol. mass) = 34.2 / 342 =0.1
Mass of water = (214.2 – 43.2) = 180 g = 0.18 kg
Molality (m) = (No. of moles of sugar / Mass of H₂O in kg) = 0.1 / 0.18 = 0.555 m, which is the required solution of the given problem. Hence, this is a correct option.

Let the final normality be N.
Total volume = (100 + 50 + 25) ml =175 ml
So, 175 * N = N1V1 + N2V2 + N3V3------(1)
Where, N1, N2, N3 are the normalities of H₂SO₄, HNO₃ and HCl respectively.
And V1, V2, V3 are the volumes of H₂SO₄, HNO₃ and HCl respectively.
Now, we have, 175 * N = ((100)(1 / 10)) + ((50)(1 / 2)) + ((25)* (1 / 5) = 40
therefore, N = 40 / 175  [where, N = normality of the mixture solution] = 0.2286, which is the required solution of the given problem, hence this is a correct option.