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