The mass of carbon anode consumed (giving only carbondioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is (Atomic mass:Al = 27)

270 kg
540 kg
90 kg
180 kg

Solution

DIRECTIONS : These are Assertion-Reason type question.Question contains two statements:Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions from the following four options.

Statement-1 :1.231 has three significant figures.
Statement-2 :All numbers right to the decimal point are significant.

Statement- 1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement -1
Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1
Statement - 1 is True,Statement- 2 is False
Statement -1 is False, Statement -2 is True

Solution

1.231 has four significant figures all no. from left to right are counted, starting with the first digit that is not zero for calculating the no. of significant figure.

DIRECTIONS : These are Assertion-Reason type question.Question contains two statements:Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions from the following four options.

Statement-1 :One mole of SO₂ contains double the number of molecules present in one mole of O₂.
Statement-2 :Molecular weight of SO₂ is double to that of O₂.

Statement- 1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement -1
Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1
Statement - 1 is True,Statement- 2 is False
Statement -1 is False, Statement -2 is True

Solution

One mole of any substance corresponding to 6.023 × 1023 entities in respective of its weight.Molecular weight of SO₂= 32 + 2 × 16 = 64gm. Molecular weight of O₂= 16 × 2 = 32gm.
∴ Molecular weight of SO₂ is double to that of O₂.

DIRECTIONS : These are Assertion-Reason type question.Question contains two statements:Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions from the following four options.

Statement-1:Volume of a gas is inversely proportional to the number of moles of gas.
Statement-2 :The ratio by volume of gaseous reactants and products is in agreement with their mole ratio.

Statement- 1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement -1
Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement - 1
Statement - 1 is True,Statement- 2 is False
Statement -1 is False, Statement -2 is True

Solution

We know that from the reaction H₂ + Cl₂ ⟶ 2HCl that the ratio of the volume of gaseous reactants and products is in agreement with their molar ratio.The ratio of H₂ : Cl₂ : HCl volume is 1: 1: 2 which is the same as their molar ratio.Thus volume of gas is directly related to the number of moles. Therefore, the assertion is false but reason is true.

In the reaction,

2Al(s) + 6HCl(aq) ⟶ 2Al³⁺(aq) + 6Cl^–(aq) + 3H₂(g)

11.2L H₂(g) at STP is produced for every mole HCl(aq)consumed
6L HCl(aq) is consumed for every 3L H₂(g) produced
33.6 L H₂(g) is produced regardless of temperature and pressure for every mole Al that reacts
67.2 H₂(g) at STP is produced for every mole Al that reacts.

Solution

The number of atoms in 0.1 mol of a triatomic gas is:(NA= 6.02 × 10²³ mol–1)

6.026 × 10²²
1.806 × 10²³
3.600 × 10²³
1.800 × 10²²

Solution

The number of atoms in 0.1 mole of a triatomic gas= 0.1 ×3 × 6.023 × 10²³= 1.806 × 10²³

10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:

3 mol
4mol
1 mol
2 mol

Solution

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C,38.71% and H,9.67%.The empirical formula of the compound would be :

CH₃O
CH₂O
CHO
CH₄O

Solution

Number of moles of MnO₄^– required to oxidize one mole of ferrous oxalate completely in acidic medium will be :

0.6 moles
0.4 moles
7.5 moles
0.2 moles

Solution

From above equation 6 moles MnO₄^-required to oxidise 10 moles of oxalate.Thus number of moles of MnO₄^-required to oxidise one mole of oxalate = \(\frac{6}{10}\) = 0.6 moles

The number of moles of KMnO₄ reduced by one mole of KI in alkaline medium is:

one
two
five
one fifth

Solution

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