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
C.
1.6 g
Mass of one mole of methane (CH4) = 16 g
Mass of 0.1 mole of methane= 16 × 0.1 g = 1.6 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
A mixture of CaCl2 and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the calcium 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 is (atomic mass of Ca = 40)
75
31.5
40.2
25