MHT-CET 2021: Some Basic Concepts of Chemistry PYQs

MCQ Quiz with Images
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What is the volume occupied by 16 g methane gas at STP?

A. 1140 cm³
B. 22400 cm³
C. 214 cm³
D. 12.4 cm³
The molar mass of methane (CH₄) is approximately 16 g/mol. Therefore, 16 g of methane is 16 g / 16 g/mol = 1 mole. At STP, one mole of any gas occupies 22.4 dm³, which is equivalent to 22400 cm³.

What is the mass of potassium chloride produced when 12.25 g potassium chlorate undergo decomposition?(At mass: K = 39, Cl = 35.5, O = 16)

A. 16.0 g
B. 14.9 g
C. 7.45 g
D. 4.25 g
The balanced chemical equation for the decomposition of potassium chlorate is 2KClO₃ → 2KCl + 3O₂. This means 2 moles of potassium chlorate produce 2 moles of potassium chloride. Since the molar mass of potassium chlorate (KClO₃) is 122.5 g/mol, 12.25 g of potassium chlorate contains 12.25 g / 122.5 g/mol = 0.1 moles. Therefore, 0.1 moles of potassium chlorate will produce 0.1 * 745 = 7.45 g of potassium chloride.

How many moles of urea are present in 5.4 g?(Molar mas = 60)

A. 2.9
B. 0.09
C. 1.2
D. 2.4
The molar mass of urea (NH₂CO) is 60 g/mol. Therefore, 5.4 g of urea contains 5.4 g / 60 g/mol = 0.09 moles.

What is the mass of 33.6 dm³ of methane gas at STP?

A. 4.8 × 10⁻² kg
B. 3.3 × 10⁻² kg
C. 1.6 × 10⁻² kg
D. 2.4 × 10⁻² kg
At STP, one mole of any gas occupies 22.4 dm³. Therefore, 33.6 dm³ of methane contains 33.6 dm³ / 22.4 dm³/mol = 1.5 moles. Since the molar mass of methane (CH₄) is approximately 16 g/mol, 1.5 moles of methane will weigh 1.5 * 16 = 24 g or 2.4 × 10² kg.

How many molecules of ammonia gas are present in 67.2 dm³, measured at STP?

A. 2.0 × 10²⁴
B. 1.0 × 10²³
C. 1.8 × 10²²
D. 5.0 × 10²⁴

What is the total number of molecules present in 224 cm³ of a gas at STP?

A. 6.022 × 10²⁰
B. 6.022 × 10²³
C. 6.022 × 10²²
D. 6.022 × 10²¹

Which of the following pair of compounds explains law of multiple proportions?

A. H₂ and O₂
B. CO and CO₂
C. N₂ and O₂
D. HClO₄ and NaClO₄
The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers. CO and CO₂ are both compounds of carbon and oxygen, and the ratio of oxygen to carbon in CO to CO₂ is different from that in CO₂, demonstrating the law of multiple proportions.

Number of molecules present in 5.4 g of urea is (Molar mass = 60 g mol⁻¹)

A. 6.0 × 10²²
B. 5.4 × 10²²
C. 9.0 × 10²²
D. 3.5 × 10²³
The molar mass of urea (NH₂CO) is 60 g/mol. Therefore, 5.4 g of urea contains 5.4 g / 60 g/mol = 0.09 moles. Since one mole contains Avogadro's number of molecules (approximately 6.022 × 10²³ molecules), 0.09 moles contain 0.09 * 6.022 × 10²³ = 5.4 × 10²² molecules.

How many atoms of Argon are present in 52 mole of it? (At. Mass of Ar = 39)

A. 1.1 × 10²³
B. 1.5 × 10²⁵
C. 3.1 × 10²⁵
D. 1.2 × 10²³
One mole contains Avogadro's number of molecules, which is approximately 6.022 × 10²³ molecules. Since each molecule of argon (Ar) is a monoatomic molecule, 52 moles of argon contain 52 * 6.022 × 10²³ = 3.1 × 10²⁵ atoms.

What is the volume occupied by 24 g methane gas at STP?

A. 33.6 dm³
B. 22.4 dm³
C. 67.2 dm³
D. 44.8 dm³
The molar mass of methane (CH₄) is approximately 16 g/mol. Therefore, 24 g of methane contains 24 g / 16 g/mol = 1.5 moles. At STP, one mole of any gas occupies 22.4 dm³, so 1.5 moles of methane will occupy 1.5 * 22.4 = 33.6 dm³.

What is the mass of 44.8 dm³ of methane gas under STP conditions?

A. 24 g
B. 32 g
C. 48 g
D. 16 g
At STP, one mole of any gas occupies 22.4 dm³. Therefore, 44.8 dm³ of methane contains 444.8 dm³ / 22.4 dm³/mol = 1.9 moles. Since the molar mass of methane (CH₄) is approximately 16 g/mol, 1.9 moles of methane will weigh 1.9 * 16 = 32 g.

Which of the following pair of compounds does not explain law of multiple proportions?

A. SO₂ and SO₃
B. O₂ and O₃
C. CO and CO₂
D. H₂O and H₂O₂
The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers. O₂ and O₃ do not demonstrate the law of multiple proportions because they are not compounds of the same elements in different proportions.

What is the SI unit of density?

A. kg dm³
B. kg m⁻³
C. kg m³
D. kg dm⁻³
The SI unit of density is kilograms per cubic meter (kg/m⁻³), which is the same as grams per cubic centimeter (g/cm⁻³).

How many molecules are present in 22400 cm³ of a gas at STP?

A. 22.4 × 10²⁰
B. 6.022 × 10²³
C. 6.022 × 10²⁰
D. 22.4 × 10²³
At STP, one mole of any gas occupies 222.4 dm³ (or 22400 cm³) and contains Avogadro's number of molecules (approximately 6.022 × 10²³ molecules). Therefore, 22400 cm³ of any gas contains 1 mole, which is equivalent to 6.022 × 10²³ molecules.

How many grams of H₂O are present in 0.25 mol of it?

A. 0.25 g
B. 5.4 g
C. 4.5 g
D. 6.1 g
The molar mass of water (H₂O) is 18 g/mol. Therefore, 0.25 mol of water contains 0.25 mol * 18 g/mol = 4.5 g.

What amount of oxygen is used at STP to obtain 9 dm³ water from sufficient amount of hydrogen gas?

A. 5.6 dm³
B. 22.4 dm³
C. 16.8 dm³
D. 11.2 dm³

"Mass can neither be created nor destroyed" is the statement of

A. Gay Lussac law of gaseous volume
B. Law of definite proportion
C. Law of conservation of mass
D. Law of multiple proportions
The statement "Mass can neither be created nor destroyed" is a restatement of the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.

Find the value of-197°C temperature in Kelvin.

A. 47 K
B. 76 K
C. 470 K
D. 760 K
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. Therefore, -197°C is equal to -197 + 273.15 = 76.15 K.

What is the volume (in dm³) occupied by 75 g ethane at STP?

A. 60.0
B. 56.0
C. 22.4
D. 44.8
The molar mass of ethane (C₂H₆) is approximately 30 g/mol. Therefore, 75 g of ethane contains 75 g / 30 g/mol = 2.5 moles. At STP, one mole of any gas occupies 22.4 dm³, so 2.5 moles of ethane will occupy 2.5 * 22.4 = 56.0 dm³.

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