RC Mukherjee Test: Laws of Chemical Combinations - NEET NCERT Based Tests

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

x + 5.85 = 8.5 + 14.35 ⇒ x = 17 g

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

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The Law of Constant Proportions, also known as the Law of Definite Proportions, states that a chemical compound always contains the same elements in the exact same proportion by mass, regardless of the size or source of the sample.

In this case, we have two different samples of iron oxide being reduced to iron:

Both experiments show a similar proportion of iron in iron oxide, confirming that iron oxide always contains iron in a constant proportion by mass. This consistency in the ratio of iron to iron oxide illustrates the Law of Constant Proportions.

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.

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.

The correct answer is B: PbO, PbO₂

The Law of Multiple Proportions states that when 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.

Let's analyze the pairs:

• A: PH₃, HCl

  • PH₃ is a compound of phosphorus and hydrogen.
  • HCl is a compound of hydrogen and chlorine.
  • These are different elements, so they do not illustrate the law of multiple proportions.

• B: PbO, PbO₂

  • Both compounds contain lead (Pb) and oxygen (O).
  • In PbO, there is one atom of oxygen for one atom of lead.
  • In PbO₂, there are two atoms of oxygen for one atom of lead.
  • The ratio of oxygen in PbO to PbO₂ is 1:2, which is a simple whole number ratio. This fits the law of multiple proportions.

• C: H₂S, SO₂

  • H₂S is a compound of hydrogen and sulfur.
  • SO₂ is a compound of sulfur and oxygen.
  • These involve different elements with sulfur, not illustrating the law of multiple proportions.

• D: CuCl₂, CuSO₄

  • CuCl₂ is a compound of copper and chlorine.
  • CuSO₄ is a compound of copper, sulfur, and oxygen.
  • These are different elements combined with copper, so they do not illustrate the law of multiple proportions.

Therefore, PbO and PbO₂ are the correct pair that illustrates the law of multiple proportions.

The correct answer is C: Gay Lussac's law of chemical combination is valid for all substances.

Explanation:

Let's analyze each statement:

• A: Gram atomic mass of an element may be defined as the mass of Avogadro's number of atoms.

  • This statement is correct. The gram atomic mass (or atomic mass in grams) is the mass of one mole (Avogadro's number) of atoms of an element.

• B: The molecular mass of a diatomic elementary gas is twice its atomic mass.

  • This statement is correct. For a diatomic gas like O₂ or N₂, the molecular mass is indeed twice the atomic mass since there are two atoms in each molecule.

• C: Gay Lussac's law of chemical combination is valid for all substances.

  • This statement is not correct. Gay Lussac's law of combining volumes applies to gases only and states that when gases react, they do so in volumes that bear a simple whole number ratio to one another and to the volumes of the products (if gaseous), provided all measurements are made under the same conditions of temperature and pressure. It does not apply to all substances, especially solids and liquids.

• D: A pure compound has always a fixed proportion of masses of its constituents.

  • This statement is correct and refers to the Law of Definite Proportions, which states that a chemical compound always contains its component elements in a fixed ratio by mass.

Therefore, statement C is the one that is not correct.

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  

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

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

(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

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

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