## The volume of 0.25 M H3PO3 required to neutralise 25 ml of 0.03 M Ca(OH)2 is

Q: The volume of 0.25 M H3PO3 required to neutralise 25 ml of 0.03 M
Ca(OH)2 is

(A) 1.32 mL

(B) 3 mL

(C) 26.4 mL

(D) 2.0 mL

Solution: Meq. of H3PO4 = Meq. of Ca(OH)2

V × 0.25 × 2 = 25 × 0.03 × 2 (H3PO3 is dibasic acid)

$\large V = \frac{25 \times 3 \times 2}{25 \times 2}$

V  = 3mL

## 1.5 litre of a solution of normality N and 2.5 litre of 2M HCl are mixed together…

Q: 1.5 litre of a solution of normality N and 2.5 litre of 2M HCl are mixed together. The resultant solution had a normality 5. The value of N is

(A) 6

(B) 10

(C) 8

(D) 4

Solution: Eq. of 1.5 litre solution + Eq. of 2.5 litre solution = Eq. of resultant of solution

1.5 × N + 2.5 × 2 = 4 × 5

N = (20-5)/1.5= 10

## 20g of sample of Ba(OH)2 is dissolved in 50 mL of 0.1 N HCl solution. The excess of HCl was titrated…

Q: 20g of sample of Ba(OH)2 is dissolved in 50 mL of 0.1 N HCl solution. The excess of HCl was titrated with 0.1 N NaOH. The volume of NaOH used was 20 ml. The percentage of Ba(OH)2 in the sample is

(A) 1.28

(B) 2.56

(C) 3.24

(D) 4.86

Solution: Equivalents of Ba(OH)2 = eqts of HCl used

$\large \frac{W}{171/2} = 3 \times 10^{-3}$

W = 0.256g

% of Ba(OH)2 = (0.256/20)100 = 1.28%

## A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water…

Q: A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.1596g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

(A) 27.9

(B) 159.6

(C) 79.8

(D) 55.8

Solution: Valency of metal in Z2O3 = 3

Z2O3 + 3H2 → 2Z + 3H2O

0.1596g of Z2O3 reacts with 6 mg of H2

1g H2 will react with g of Z2O3

equivalent weight of Z2O3 = 0.1596/0.006 = 26.6g

equivalent weight of Z2O3 = 26.6

= eq.wt. of Z + eq. wt. of = E + 8

E + 8 = 26.6

E = 18.6

Atomic weight of Z = 18.6 × 3 = 55.8

## When 0.1 mole of H3BO3 is titrated against NaOH of 1 M using phenolphthalein indicator

Q: When 0.1 mole of H3BO3 is titrated against NaOH of 1 M using phenolphthalein indicator what volume of NaOH is used up to the end point?

(A) 300 ml

(B) 100 ml

(C) 3000 ml

(D) 1000 ml

Solution: H3BO3 is a monobasic acid

equivalents of H3BO3 = equivalents of NaOH

0.1 = 1 × V

V = 0.1 litre or 100 ml