1 c.c. N₂O at NTP contains :

\(\frac{1.8}{224}\) × 10²² atoms
\(\frac{6.02}{22400}\) × 10²³ atoms
\(\frac{1.32}{224}\) × 10²³ atoms
All these

Solution

At NTP 22400 cc of N₂O = 6.02 × 10²³ molecules

1 cc N₂O = \(\frac{6.02 \times 10^{23}}{22400}\) molecules
= \(\frac{3 \times 6.02 \times 10^{23}}{22400}\) atoms = \(\frac{1.8}{224}\) × 10²² atoms

No. of electrons in a molecule of NO₂= 7 + 7 + 8= 22
Hence no. of electrons = \(\frac{6.02 \times 10^{23}}{22400}\) × 22 electrons = \(\frac{1.32 \times 10^{23}}{224}\)

6.8 gm H₂O₂ present in 100 ml of its solution. What is the molarity of solution?

1 M
2 M
3 M
0.5 M

Solution

∵ Weight of H₂O₂ in 100 ml of H₂O₂ solution = 6.8 gm
∵ Weight of H₂O₂ in 1000 ml of its solution = 6.8×10 = 68g Molecular weight of H₂O₂ = 34
Then, Molarity = \(\frac{68}{34}\) = 2M

The volume of chlorine at STP required to liberate all the bromine and iodine in 100 ml of 0.1 Meach of KI and MBr₂ will be:

0.224 L
0.336 L
0.448 L
0.560 L

Solution

On reduction 1.644 gm of hot iron oxide give 1.15 gm of iron.Evaluate the equivalent weight of iron.

18.62
19.13
18.95
12.95

Solution

Weight of iron oxide = 1.644 gm
Weight of iron after reduction = 1.15 gm
weight of displaced oxygen = 1.644– 1.15 = 0.494 gm
Equivalent weight of iron = \(\frac{1.15}{0.494}\) × 8 = 18.62
Thus equivalent weight of metal is = 18.62.

If 0.5 mol of BaCl₂ is mixed with 0.2 mole of Na₃PO₄, find the maximum amount of Ba₃(PO₄)₂ that can be formed.

1 mole
0.5 mole
0.1 mole
0.01 mole

Solution

10 moles SO₂ and 15 moles O₂ were allowed to react over a suitable catalyst. 8 moles of SO₃ were formed. The remaining moles of SO₂ and O₂ respectively are –

2 moles, 11 moles
2 moles, 8 moles
4 moles, 5 moles
8 moles, 2 moles

Solution

On adding excess of CaCl₂ to a solution containing Na₂CO₃ and NaHCO₃, x g of precipitate was obtained. On adding in drops to the filtrate, a further yg of precipitate was obtained.In another experiment to the same amount of solution excess of CaCl₂ was added, boiled and filtered. The amount of the precipitate in the second experiment would be

x+y
x+^y/₂
\(\frac{x+y}{2}\)
noneof these

Solution

The hydrated salt Na₂CO₃ x H₂O₃ undergoes 63% loss in mass on heating and becomes anhydrous. The value of x is

Solution

The loss in mass is due to elimination of water of crystallisation of Na₂CO₃ x H₂O₃

Hence,\(\frac{18x \times 6}{106 + 18x}\) = 63 ⇨x=10

12 L of H₂ and 11.2 L of Cl₂ are mixed and exploded. Find the composition by volume of mixture.

11.2, 11.2, 22.4
0.8, 0, 22.4
0.8, 0.8, 22.4
0.8, 11.2, 22.4

Solution

0.5400 g of a metal X yields 1.020 g of its oxide X₂O₃.The number of moles of X is :

0.01
0.02
0.04
0.05

Solution

Mass of oxygen combined with 0.5400 g of X= 1.0200 – 0.5400 = 0.4800 g
Mol of X = \(\frac{2 \times 0.48}{48}\) = 0.02

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