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
D.
5 g of hydrogen
Number of moles of oxygen =
Number of atoms of oxygen = = 5 × N0 ×2
Number of moles of 5g of hydrogen =
Number of atoms in 5g of hydrogen =5 × N0 ×2
Hence, the number of atoms in 80g of oxygen is equal to the number of atoms in 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
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
The rms velocity of hydrogen is times the rms velocity of nitrogen. If T is the temperature of the gas, which of the following is true?
50 cm3 of 0.2 NHCl is titrated against 0.1 N NaOH solution. The titration is 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
25 cm3
10.5 cm3
1 g of silver gets distributed between 10 cm3 of molten zinc and 100 cm3 of molten lead at 800°C. The percentage of silver in the zinc layer is approximately
89
91
97
94