Solutions and pH Chemistry

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.

Mol. mass of CaCO₃ = 100 Mol. mass of Ca = 40 20 mg of CaCO₃ has ((40*20) / 100) = 8 mg of Ca⁺² ions Since, 8 mg of Ca⁺² in 100 L sample of hard water. 10*1000 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.

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.

In order to sediment free ribosomes, centrifugation requires 3,00,000 g.

Optimum pH of pepsin ~ 2

When a quantity like 'five millilitres' is treated as a single amount or measurement, it takes a singular verb. 'Seems' agrees with the singular idea of the amount, not the plural noun. 'Seem' would be incorrect here because we're focusing on the total quantity as one unit.

N₁V₁(Initial) = N₂V₂ (After dilution) 10 × 10 = 0.1 × V₂ or V₂ = 1000 ml So, volume of water needed = 1000 – 10 = 990 ml

N₁ V₁ (NaOH) = N₂ V₂ (HCl) 1 N x V₁ = 0.5 N x 10 ml V₁ = 5 ml

Our body works in the pH range of 7.0 - 7.5.

The isoelectric pH (pI) of leucine is approximately 6.0. This is calculated as the average of its carboxyl group pKa (~2.4) and amino group pKa (~9.6), giving pI = (2.4 + 9.6)/2 = 6.0. At this pH, leucine exists as a zwitterion with no net charge.

acidic