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Q. If 'WATER' is coded as 'XBTFS', how is 'FIRE' coded? (2021)

A. GJTF
B. GJTE
C. GJTD
D. GJTC

Solution

Each letter is shifted by +1. F->G, I->J, R->S, E->F. So, FIRE = GJTF.

Correct Answer: A — GJTF

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Q. If 'x' in base-7 is equal to 50 in decimal, what is 'x'?

A. 101
B. 110
C. 120
D. 130

Solution

50 in decimal is '110' in base-7 (1*7^2 + 1*7^1 + 0*7^0 = 50).

Correct Answer: B — 110

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Q. If 'x' is a number in base 5 represented as '320', what is its decimal equivalent?

A. 75
B. 80
C. 85
D. 90

Solution

'320' in base 5 = 3*5^2 + 2*5^1 + 0*5^0 = 75.

Correct Answer: A — 75

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Q. If 'X' is a square and 'Y' is a rectangle, which of the following is true?

A. All squares are rectangles.
B. All rectangles are squares.
C. X is not a rectangle.
D. Y is a square.

Solution

All squares are rectangles because they meet the definition of a rectangle.

Correct Answer: A — All squares are rectangles.

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Q. If 'XYZ' in a base-6 system equals 215 in decimal, what is the value of 'X'?

A. 3
B. 4
C. 5
D. 2

Solution

215 in decimal is '543' in base-6, so 'X' is 4.

Correct Answer: B — 4

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Q. If 'XYZ' in base 5 equals 100 in decimal, what is the value of 'X'?

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

Solution

'XYZ' in base 5 means X*25 + Y*5 + Z = 100. The only valid solution is X=2.

Correct Answer: B — 2

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Q. If 'XYZ' in base-3 equals 26 in decimal, what is the value of 'X'?

A. 2
B. 1
C. 0
D. 3

Solution

In base-3, 'XYZ' means X*9 + Y*3 + Z = 26. The only valid digit for X is 2.

Correct Answer: A — 2

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Q. If 'XYZ' in base-4 equals 27 in decimal, what is the value of 'X'?

A. 1
B. 2
C. 3
D. 0

Solution

In base-4, 'XYZ' = 4^2*X + 4^1*Y + 4^0*Z = 27. The only valid combination is X=3, Y=3, Z=3.

Correct Answer: C — 3

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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 0.1 M NaOH is mixed with 0.1 M HCl, what will be the resulting pH?

A. 0
B. 7
C. 14
D. 1

Solution

The reaction between NaOH and HCl neutralizes each other, resulting in a neutral solution with a pH of 7.

Correct Answer: B — 7

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Q. If 0.1 M of a strong acid is mixed with 0.1 M of a strong base, what will be the resulting pH?

A. 0
B. 7
C. 14
D. 1

Solution

Strong acid and strong base neutralize each other, resulting in a neutral solution with pH = 7.

Correct Answer: B — 7

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Q. If 0.1 M of a weak acid has a pH of 4.0, what is the Ka of the acid?

A. 1 x 10^-4
B. 1 x 10^-5
C. 1 x 10^-6
D. 1 x 10^-7

Solution

Using the formula Ka = [H+]^2 / [HA], where [H+] = 10^(-4) M and [HA] = 0.1 M, we find Ka = (10^-4)^2 / 0.1 = 1 x 10^-5.

Correct Answer: B — 1 x 10^-5

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Q. If 0.5 mol of a non-volatile solute is dissolved in 1 kg of water, what is the vapor pressure lowering? (Vapor pressure of pure water = 23.76 mmHg)

A. 1.88 mmHg
B. 2.88 mmHg
C. 3.88 mmHg
D. 4.88 mmHg

Solution

Vapor pressure lowering = (n_solute / n_solvent) * P°_solvent = (0.5 / 55.5) * 23.76 = 1.88 mmHg

Correct Answer: A — 1.88 mmHg

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Q. If 0.5 moles of a gas occupy 11.2 liters at STP, what is the molar volume of the gas?

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

Solution

At STP, 1 mole of gas occupies 22.4 L. Therefore, 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 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 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.52 °C kg/mol) (2021)

A. 0.26 °C
B. 0.52 °C
C. 1.04 °C
D. 0.78 °C

Solution

Boiling point elevation = i * Kb * m = 1 * 0.52 * 0.5 = 0.26 °C, so change = 0.52 °C.

Correct Answer: C — 1.04 °C

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Q. If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2021)

A. 0.5 P0
B. 0.25 P0
C. 0.75 P0
D. 0.1 P0

Solution

Vapor pressure lowering = X_solute * P0 = (0.5 / (0.5 + 55.5)) * P0 ≈ 0.25 P0.

Correct Answer: B — 0.25 P0

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Q. If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2020)

A. 0.5 P0
B. 0.25 P0
C. 0.75 P0
D. 0.1 P0

Solution

Vapor pressure lowering = (moles of solute / moles of solvent) * P0 = (0.5 / 55.5) * P0 ≈ 0.25 P0.

Correct Answer: B — 0.25 P0

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Q. If 0.5 moles of NaCl are dissolved in 1 liter of water, what is the concentration of NaCl in the solution?

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

Solution

Concentration (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.

Correct Answer: A — 0.5 M

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Q. If 0.5 moles of NaCl are dissolved in 1 liter of water, what is the concentration of NaCl?

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

Solution

Concentration (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.

Correct Answer: A — 0.5 M

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Q. If 0.5 moles of NaCl are dissolved in 1 liter of water, what is the molarity of the solution?

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

Solution

Molarity (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.

Correct Answer: A — 0.5 M

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Q. If 0.5 moles of NaCl is dissolved in 1 liter of water, what is the molality of the solution? (2022)

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

Solution

Molality (m) = moles of solute / kg of solvent. Assuming 1 L of water = 1 kg, molality = 0.5 moles / 1 kg = 0.5 m.

Correct Answer: A — 0.5 m

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Q. If 1 kg of water is heated from 20°C to 100°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C)

A. 3360 J
B. 4000 J
C. 4200 J
D. 4800 J

Solution

Q = m*c*ΔT = 1000 g * 4.2 J/g°C * (100°C - 20°C) = 4200 J.

Correct Answer: C — 4200 J

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Q. If 1 kg of water is heated from 25°C to 75°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C) (2021)

A. 21000 J
B. 42000 J
C. 84000 J
D. 105000 J

Solution

Q = mcΔT = (1000 g)(4.2 J/g°C)(50°C) = 210000 J.

Correct Answer: B — 42000 J

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Q. If 1 kg of water is mixed with 1 kg of sugar, what is the expected change in boiling point? (2023)

A. No change
B. Increase
C. Decrease
D. Depends on the sugar type

Solution

The addition of sugar (a non-volatile solute) will raise the boiling point of water due to boiling point elevation.

Correct Answer: B — Increase

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Q. If 1 L of a 2 M solution is diluted to 3 L, what is the new molarity of the solution?

A. 0.67 M
B. 1 M
C. 1.5 M
D. 2 M

Solution

Using the dilution formula M1V1 = M2V2, we have 2 M * 1 L = M2 * 3 L, thus M2 = 2/3 = 0.67 M.

Correct Answer: A — 0.67 M

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Q. If 1 L of a 3 M solution is diluted to 2 L, what is the new molarity?

A. 1.5 M
B. 3 M
C. 6 M
D. 0.5 M

Solution

Using the dilution formula M1V1 = M2V2, we have 3 M × 1 L = M2 × 2 L. Thus, M2 = 1.5 M.

Correct Answer: A — 1.5 M

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Q. If 1 liter of a 2 M solution is diluted to 3 liters, what is the new molarity?

A. 0.67 M
B. 1 M
C. 1.5 M
D. 2 M

Solution

Using the dilution formula M1V1 = M2V2, (2 M)(1 L) = M2(3 L) => M2 = 2/3 = 0.67 M.

Correct Answer: A — 0.67 M

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Major Competitive Exams

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