Chemistry Study Resources for Competitive Exams

Chemistry

Alcohols & Phenols Aldehydes & Ketones Amines Atomic Structure Biomolecules Carboxylic Acids Chemical Bonding Chemical Equilibrium Chemical Kinetics Coordination Compounds d & f-block Elements Electrochemistry Haloalkanes Hydrocarbons Ionic Equilibrium Mole Concept Organic Chemistry Basics p-Block Elements Periodic Table Polymers Redox Reactions Solutions States of Matter Surface Chemistry Thermodynamics

Q. How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (2023)

A. 14.9 g
B. 7.45 g
C. 29.8 g
D. 3.73 g

Solution

Molarity = moles/volume(L). Moles = 0.5 mol/L * 0.5 L = 0.25 mol. Mass = moles * molar mass = 0.25 mol * 74.5 g/mol = 18.625 g.

Correct Answer: B — 7.45 g

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Q. How many grams of KCl are needed to prepare 250 mL of a 0.5 M solution? (2023) 2023

A. 7.45 g
B. 12.5 g
C. 9.25 g
D. 5.25 g

Solution

Molar mass of KCl = 74.55 g/mol. Grams = Molarity * Volume (L) * Molar mass = 0.5 M * 0.25 L * 74.55 g/mol = 9.31 g.

Correct Answer: A — 7.45 g

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Q. How many grams of KCl are needed to prepare 250 mL of a 2 M solution? (2023)

A. 37.25 g
B. 50 g
C. 75 g
D. 25 g

Solution

Molar mass of KCl = 74.55 g/mol. Grams = moles × molar mass = 0.5 moles × 74.55 g/mol = 37.25 g.

Correct Answer: A — 37.25 g

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Q. How many grams of KCl are needed to prepare 250 mL of a 2 M solution? (Molar mass of KCl = 74.5 g/mol) (2023)

A. 37.25 g
B. 74.5 g
C. 18.625 g
D. 9.25 g

Solution

Grams = moles × molar mass = (2 moles/L × 0.25 L) × 74.5 g/mol = 37.25 g.

Correct Answer: A — 37.25 g

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Q. How many liters of a 0.5 M NaOH solution are needed to obtain 1 mole of NaOH? (2023)

A. 1 L
B. 2 L
C. 0.5 L
D. 0.25 L

Solution

Volume = moles / molarity = 1 mole / 0.5 M = 2 L.

Correct Answer: A — 1 L

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Q. How many liters of a 0.5 M solution contain 2 moles of solute? (2023)

A. 2 L
B. 4 L
C. 1 L
D. 0.5 L

Solution

Volume = moles / molarity = 2 moles / 0.5 M = 4 L

Correct Answer: B — 4 L

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Q. How many liters of a 3 M solution can be made from 6 moles of solute? (2023)

A. 1 L
B. 2 L
C. 3 L
D. 4 L

Solution

Volume = moles / molarity = 6 moles / 3 M = 2 L.

Correct Answer: B — 2 L

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Q. How many molecules are in 0.1 moles of oxygen gas (O2)? (2021)

A. 6.022 x 10^22
B. 6.022 x 10^23
C. 1.2044 x 10^23
D. 3.011 x 10^23

Solution

Number of molecules = moles x Avogadro's number = 0.1 moles x 6.022 x 10^23 molecules/mole = 6.022 x 10^22 molecules.

Correct Answer: A — 6.022 x 10^22

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Q. How many molecules are in 0.25 moles of oxygen gas (O2)? (2022)

A. 3.011 x 10^23
B. 6.022 x 10^22
C. 1.2044 x 10^23
D. 1.5055 x 10^23

Solution

Number of molecules = moles x Avogadro's number = 0.25 moles x 6.022 x 10^23 molecules/mole = 1.5055 x 10^23 molecules.

Correct Answer: B — 6.022 x 10^22

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Q. How many molecules are present in 2 moles of water (H2O)? (2022) 2022

A. 6.022 x 10^23
B. 1.2044 x 10^24
C. 3.011 x 10^23
D. 1.2044 x 10^23

Solution

Number of molecules = moles x Avogadro's number = 2 moles x 6.022 x 10^23 molecules/mole = 1.2044 x 10^24 molecules.

Correct Answer: B — 1.2044 x 10^24

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Q. How many moles are in 100 grams of NaOH? (2022)

A. 2.5
B. 1.25
C. 0.5
D. 3

Solution

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Number of moles = mass/molar mass = 100 g / 40 g/mol = 2.5 moles.

Correct Answer: B — 1.25

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Q. How many moles are in 32 grams of O2? (2021)

A. 1
B. 2
C. 0.5
D. 4

Solution

Molar mass of O2 = 16*2 = 32 g/mol. Number of moles = mass/molar mass = 32g/32g/mol = 1 mole.

Correct Answer: A — 1

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Q. How many moles of solute are present in 500 mL of a 0.2 M solution? (2021) 2021

A. 0.1 moles
B. 0.5 moles
C. 0.2 moles
D. 0.25 moles

Solution

Moles = Molarity * Volume (L) = 0.2 M * 0.5 L = 0.1 moles.

Correct Answer: A — 0.1 moles

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Q. How many orbitals are present in the d-subshell? (2023)

A. 1
B. 3
C. 5
D. 7

Solution

The d-subshell has 5 orbitals.

Correct Answer: C — 5

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Q. How many orbitals are present in the p subshell? (2023)

A. 1
B. 2
C. 3
D. 5

Solution

The p subshell has 3 orbitals (px, py, pz).

Correct Answer: C — 3

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Q. Identify the oxidizing agent in the reaction: 2H2 + O2 → 2H2O. (2023)

A. H2
B. O2
C. H2O
D. None of the above

Solution

O2 is the oxidizing agent as it gains electrons and is reduced to H2O.

Correct Answer: B — O2

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Q. If 0.1 M acetic acid (CH3COOH) is 1% ionized, what is the concentration of H+ ions?

A. 0.001 M
B. 0.01 M
C. 0.1 M
D. 0.2 M

Solution

Ionization = 1% of 0.1 M = 0.001 M, so [H+] = 0.001 M.

Correct Answer: B — 0.01 M

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Q. If 0.1 M acetic acid has a pH of 2.87, what is the concentration of H+ ions?

A. 0.001 M
B. 0.01 M
C. 0.1 M
D. 0.5 M

Solution

[H+] = 10^(-pH) = 10^(-2.87) ≈ 0.001 M.

Correct Answer: B — 0.01 M

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Q. If 0.5 moles of a gas occupy 12 liters at STP, what is the molar volume of the gas? (2023)

A. 22.4 L
B. 24 L
C. 12 L
D. 6 L

Solution

At STP, 1 mole of any gas occupies 22.4 L. Therefore, 0.5 moles occupy 0.5 x 22.4 L = 11.2 L, confirming the molar volume is 22.4 L.

Correct Answer: A — 22.4 L

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Q. If 0.5 moles of a gas occupy 12.2 liters at STP, what is the molar volume of the gas? (2023)

A. 22.4 L
B. 24 L
C. 12.2 L
D. 6.1 L

Solution

At STP, 1 mole of gas occupies 22.4 liters. Therefore, the molar volume is 22.4 L.

Correct Answer: A — 22.4 L

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Q. If 1 mole of a gas occupies 22.4 liters at STP, how many liters will 0.5 moles occupy? (2023)

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 5.6 L

Solution

At STP, 1 mole of gas occupies 22.4 L. Therefore, 0.5 moles will occupy 0.5 x 22.4 L = 11.2 L.

Correct Answer: A — 11.2 L

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Q. If 10 g of glucose (C6H12O6) is dissolved in 200 g of water, what is the mass percentage of the solution? (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: B — 5.00%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the solution? (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 20.00%

Solution

Mass percent = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) × 100 = 4.76%.

Correct Answer: B — 5.00%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: B — 5.00%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the sugar solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the sugar solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) × 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of sugar in the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / total mass of solution) x 100. Total mass = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the sugar solution? (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar is dissolved in 100 g of water, what is the mass percent of sugar in the solution? (2023)

A. 10%
B. 5%
C. 20%
D. 15%

Solution

Mass percent = (mass of solute / total mass of solution) x 100. Total mass = 10 g + 100 g = 110 g. Mass percent = (10 g / 110 g) x 100 = 9.09%, approximately 10%.

Correct Answer: A — 10%

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Chemistry

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