The volume of 10 N and 4 N HCl required to make 1 L of 7 N HCl are

• 0.50 L of 10 N HCl and 0.50 L of 4 N HCl

• 0.60 L of 10 N HCl and 0.40 L of 4 N HCl

• 0.80 L of 10 N HCl and 0.20 L of 4 N HCl

• 0.75 L of 10 N HCl and 0.25 L of 4 N HCl

An oxide of the element contains 20% O₂ by weight. Calculate the equivalent weight of the element.

• 8

• 16

• 32

• 12

For the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, the volume of carbon monoxide required to reduce one mole of ferric oxide is

• 22.4 dm³

• 44.8 dm³

• 67.2 dm³

• 11.2 dm³

10 cm³ of 0.1 N monobasic acid requires 15 cm³ of sodium hydroxide solution whose normality is

• 1.5 N

• 0.15 N

• 0.066 N

• 0.66 N

Mass of 0.1 mole of methane is

• 1 g

• 16 g

• 1.6 g

• 0.1 g

80 g of oxygen contains as many atoms as in

• 80 g of hydrogen

• 1 g of hydrogen

• 10 g of hydrogen

• 5 g of hydrogen

Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (Atomic mass of calcium = 40)

• 300 cm³

• 200 cm³

• 500 cm³

• 400 cm³

A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate is

• 182

• 168

• 192

• 188

A mixture of CaCl₂ and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the Ca²⁺ ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of Cao. The percentage of NaCl in the mixture (atomic mass of Ca = 40) is

• 75

• 30.6

• 25

• 69.4

160.

50 cm³ of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration was discontinued after adding 50 cm³ of NaOH. The remaining titration is completed by adding 0.5 N KOH. The volume of KOH required for completing the titration is 

• 12 cm³

• 10 cm³

• 21.0 cm³

• 16.2 cm³