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RC Mukherjee Test: Laws of Chemical Combinations


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RC Mukherjee Test: Laws of Chemical Combinations - Question 1

4.88 g of KCIO3 when heated produced 1.92 g of O2 and 2.96 gof KCl. Which ofthe following statements regarding the experiment is correct?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 1


Since, mass of the products (2.96 + 1.92) is equal to mass of the reactant, this illustrates the law of conservation of mass.

RC Mukherjee Test: Laws of Chemical Combinations - Question 2

How much mass of silver nitrates will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrates if law of conservation of mass is followed?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 2


x + 5.85 = 8.5 + 14.35 ⇒ x = 17 g

RC Mukherjee Test: Laws of Chemical Combinations - Question 3

What mass of hydrochloric acid is needed to decompose 50 g of limestone?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 3


100 g of CaCO3 requires 73 g of HCl
∴ 50 g of CaCO3 requires 73/100 x 50 = 36.5 g of HCl

RC Mukherjee Test: Laws of Chemical Combinations - Question 4

What mass of sodium chloride would be decomposed by 9.8 g of sulphuric acid if 12 g of sodium bisulphate and 2.75 g of hydrogen chloride were produced in a reaction?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 4


x + 9.8 = 12 + 2.75 ⇒ x = 4.95g

RC Mukherjee Test: Laws of Chemical Combinations - Question 5

In an experiment, 2.4g of iron oxide on reduction with hydrogen gave 1.68 g of iron. In another experiment, 2.9 g of iron oxide gave 2.09 g of iron on reduction.Which law is illustrated from the above data?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 5

In first experiment,
2.4 g of iron oxide gave 1.68 g of iron.
Mass of iron oxide = 2.4 g
Mass of iron = 1.68 g
Mass of oxygen = 2.4 - 1.68 = 0.72
Ratio of masses of iron and oxygen = 1.68/0.72 = 7:3
In second experiment,
Mass of iron oxide = 2.9 g
Mass of iron = 2.09 g
Mass of oxygen = 2.9 - 2.09 = 0.81
Ratio of masses of iron and oxygen = 2.09/0.81 = 7.3
The same ratio confirms that these experiments clarify Law of constant proportions.

RC Mukherjee Test: Laws of Chemical Combinations - Question 6

The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

Which law of chemical combination is obeyed by the above experimental data?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 6

According to the law of multiple proportions, "if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers."
Fixing the mass of dinitrogen as 28g, masses of dioxygen combined will be 32, 64, 32 and 96 in the given figures. They are in the ratio of 1 : 2 : 1 : 3.

RC Mukherjee Test: Laws of Chemical Combinations - Question 7

Which of the following statements illustrates the law of multiple proportions?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 7

In XO, 50 g of element combines with 50 g of oxygen.
∴ 1 g of element combines with 1 g of oxygen.
In XO2,40 g of element combines with 60 g of oxygen.
∴ l.g of element combines with 1.5 g of oxygen.
Thus, ratio of masses of oxygen which combines with 1 g of element is 1:1.5 or 2:3. This is in accordance with the law of multiple proportions: In (b) the law of reciprocal proportions is followed. In (c) law of conservation of mass is followed while in (d) Avogadro's law is followed.

RC Mukherjee Test: Laws of Chemical Combinations - Question 8

Which of the following pairs illustrates the law of multiple proportions?

RC Mukherjee Test: Laws of Chemical Combinations - Question 9

Given below are few statements. Mark the statement which is not correct.

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 9

Gay Lussac's law is true only for gaseous substances.

RC Mukherjee Test: Laws of Chemical Combinations - Question 10

What quantity of copper oxide will react with 2.80 L of hydrogen at NTP?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 10


22.4 L of H2 = 79.5g of CuO
2.80 L of H2 = 79.5/22.4 x 2.80 = 9.9 g of CuO  

RC Mukherjee Test: Laws of Chemical Combinations - Question 11

At NTP, 1L of O2 reacts with 3 L of carbon monoxide. What will be the volume of CO and CO2 after the reaction?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 11


1 vol of CO2 reacts with 2 vol of CO
1 L of CO2 reacts with 2 L of CO
CO left after reaction = 3 - 2 = 11
1 L of O2 produces 2 L of CO2.
Hence, after the reaction, CO = 1 L, CO2 = 2 L

RC Mukherjee Test: Laws of Chemical Combinations - Question 12

Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide. How much volume of CO2 will be obtained by thermal decomposition of 50 g of CaCO3?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 12


100 g of CaCO3 at STP gives 22.4 L of CO2
50 g of CaCO3 will produce 22.4/100 x 50 = 11.2 L of CO2

RC Mukherjee Test: Laws of Chemical Combinations - Question 13

Chlorine gas is prepared by reaction of H2SO4 with MnO2 and NaCl. What volume of CI2 will be produced at STP if 50 g of NaCl is taken in the reaction?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 13


(2 x 58.5 = 117 g) 22.4 L (STP)
117g of NaCl = 22.4 L of Cl2
50 g of NaCl = 22.4/117 x 50 = 9.57 L of Cl2 at STP

RC Mukherjee Test: Laws of Chemical Combinations - Question 14

HCl is produced in the stomach which can be neutralised by Mg(OH)2 in the form of milk of magnesia. How much Mg(OH)2 is required to neutralise one mole of stomach acid?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 14

Mg(OH)2 + 2HCl → MgCl2 + 2H2O
No. of moles of Mg(OH)2 required for 2 moles of HCl = 1
No. of moles of Mg(OH)2 required for 1 mole of HCl = 0.5
Mass of 0.5 mol of Mg(OH)2 = 58.33 x 0.5 = 29.16 g

RC Mukherjee Test: Laws of Chemical Combinations - Question 15

Iron can be obtained by reduction, of iron oxide (FC3O4) with CO according to the reaction:
Fe3O4 + 4CO → 3Fe + 4CO2
How many kg of Fe3O4 should be heated with CO to get 3 kg of iron?

Detailed Solution for RC Mukherjee Test: Laws of Chemical Combinations - Question 15


3 moles of Fe is produced from 1 mole of Fe3O4
168 g of Fe is producedfrom 232 g of Fe3O4.
3 kg of Fe will be produced from 232/168 x 3000 g
= 4142.8 g.or 4.14 kg of Fe3O4

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