Q. If 10 grams of NaCl is dissolved in enough water to make 500 mL of solution, what is the molality of the solution? (Molar mass of NaCl = 58.5 g/mol)
A.
0.34 m
B.
0.17 m
C.
0.85 m
D.
0.50 m
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Solution
Moles of NaCl = 10 g / 58.5 g/mol = 0.171 moles. Mass of solvent (water) = 0.5 kg. Molality (m) = moles of solute / kg of solvent = 0.171 moles / 0.5 kg = 0.34 m.
Correct Answer: A — 0.34 m
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Q. If 10 grams of NaOH are dissolved in water, how many moles of NaOH are present? (Molar mass of NaOH = 40 g/mol)
A.
0.25 moles
B.
0.5 moles
C.
1 mole
D.
2.5 moles
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Solution
To find the number of moles, use the formula: moles = mass/molar mass. Thus, 10 g / 40 g/mol = 0.25 moles.
Correct Answer: B — 0.5 moles
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Q. If 10 grams of NaOH is dissolved in 500 mL of solution, what is the molality of the solution? (Molar mass of NaOH = 40 g/mol)
A.
0.5 m
B.
1 m
C.
2 m
D.
0.25 m
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Solution
Moles of NaOH = 10 g / 40 g/mol = 0.25 moles. Mass of solvent (water) = 0.5 kg. Molality (m) = moles of solute / kg of solvent = 0.25 moles / 0.5 kg = 0.5 m.
Correct Answer: B — 1 m
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Q. If 10 grams of NaOH is dissolved in enough water to make 500 mL of solution, what is the molarity of the solution? (Molar mass of NaOH = 40 g/mol)
A.
0.5 M
B.
1 M
C.
2 M
D.
0.25 M
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Solution
Moles of NaOH = 10 g / 40 g/mol = 0.25 moles. Molarity = 0.25 moles / 0.5 L = 0.5 M.
Correct Answer: B — 1 M
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Q. If 10 liters of a solution contains 30% salt, how much salt is present in the solution?
A.
3 liters
B.
2 liters
C.
4 liters
D.
1 liter
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Solution
30% of 10 liters = 0.3 * 10 = 3 liters of salt.
Correct Answer: A — 3 liters
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Q. If 100 g of glucose (C6H12O6) is dissolved in 1 L of solution, what is the molarity of the solution? (Molar mass of glucose = 180 g/mol)
A.
0.56 M
B.
1.0 M
C.
0.33 M
D.
0.75 M
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Solution
Moles of glucose = 100 g / 180 g/mol = 0.556 moles. Molarity = moles of solute / liters of solution = 0.556 moles / 1 L = 0.56 M.
Correct Answer: A — 0.56 M
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Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature?
A.
50°C
B.
25°C
C.
75°C
D.
0°C
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Solution
The final temperature will be 50°C due to equal masses and specific heat capacities.
Correct Answer: A — 50°C
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Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature of the mixture?
A.
50°C
B.
25°C
C.
75°C
D.
0°C
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Solution
Using the principle of conservation of energy, the final temperature will be 50°C.
Correct Answer: A — 50°C
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Q. If 100 g of water at 80°C is mixed with 200 g of water at 20°C, what will be the final temperature?
A.
30°C
B.
40°C
C.
50°C
D.
60°C
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Solution
Using the principle of conservation of energy, the final temperature can be calculated to be 40°C.
Correct Answer: B — 40°C
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Q. If 100 J of heat is added to a system and 40 J of work is done by the system, what is the change in internal energy?
A.
60 J
B.
40 J
C.
100 J
D.
140 J
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Solution
According to the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 100 J - 40 J = 60 J.
Correct Answer: A — 60 J
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Q. If 100 J of heat is added to a system at a constant temperature of 300 K, what is the change in entropy?
A.
0.33 J/K
B.
0.25 J/K
C.
0.5 J/K
D.
0.75 J/K
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Solution
The change in entropy ΔS = Q/T = 100 J / 300 K = 0.33 J/K.
Correct Answer: A — 0.33 J/K
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Q. If 1000 J of heat is added to a gas and it expands doing 400 J of work, what is the change in internal energy? (2023)
A.
600 J
B.
400 J
C.
1000 J
D.
200 J
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Solution
Using the first law of thermodynamics: ΔU = Q - W = 1000 J - 400 J = 600 J.
Correct Answer: A — 600 J
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Q. If 1000 J of heat is added to a system and it does 400 J of work, what is the change in internal energy? (2021)
A.
600 J
B.
400 J
C.
1000 J
D.
1400 J
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Solution
Using the first law of thermodynamics, ΔU = Q - W = 1000 J - 400 J = 600 J.
Correct Answer: A — 600 J
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Q. If 10^(2x) = 100, what is the value of x?
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Solution
100 = 10^2, so 2x = 2, hence x = 1.
Correct Answer: B — 0.5
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Q. If 10^(x) = 1000, what is the value of x? (2023)
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Solution
Since 1000 can be expressed as 10^3, we have 10^x = 10^3, thus x = 3.
Correct Answer: B — 3
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Q. If 10^(x+2) = 1000, what is the value of x? (2023)
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Solution
Since 1000 can be expressed as 10^3, we have 10^(x+2) = 10^3, thus x + 2 = 3, leading to x = 1.
Correct Answer: A — 1
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Q. If 10^x = 0.01, what is the value of x?
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Solution
Since 0.01 = 10^(-2), we have x = -2.
Correct Answer: A — -2
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Q. If 15 g of sugar is dissolved in 300 mL of water, what is the concentration in g/mL? (2020)
A.
0.05 g/mL
B.
0.1 g/mL
C.
0.15 g/mL
D.
0.2 g/mL
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Solution
Concentration = mass / volume = 15 g / 300 mL = 0.05 g/mL.
Correct Answer: B — 0.1 g/mL
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Q. If 15 grams of HCl is dissolved in 500 mL of solution, what is the molarity of the solution? (Molar mass of HCl = 36.5 g/mol)
A.
0.82 M
B.
1.0 M
C.
0.5 M
D.
1.5 M
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Solution
Moles of HCl = 15 g / 36.5 g/mol = 0.41096 moles. Molarity = 0.41096 moles / 0.5 L = 0.82192 M.
Correct Answer: A — 0.82 M
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Q. If 15 grams of sugar is dissolved in 250 mL of solution, what is the molarity of the solution? (Molar mass of sugar = 180 g/mol)
A.
0.33 M
B.
0.5 M
C.
0.25 M
D.
0.75 M
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Solution
Moles of sugar = 15 g / 180 g/mol = 0.0833 moles. Molarity = 0.0833 moles / 0.25 L = 0.333 M.
Correct Answer: A — 0.33 M
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Q. If 15 grams of sugar is dissolved in 250 mL of water, what is the molarity of the solution? (Molar mass of sugar = 180 g/mol)
A.
0.33 M
B.
0.5 M
C.
0.25 M
D.
0.75 M
Show solution
Solution
Moles of sugar = 15 g / 180 g/mol = 0.0833 moles. Molarity = 0.0833 moles / 0.25 L = 0.333 M.
Correct Answer: A — 0.33 M
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Q. If 2 moles of a gas occupy 44.8 L at STP, what is the molar volume of the gas?
A.
22.4 L
B.
44.8 L
C.
11.2 L
D.
33.6 L
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Solution
At STP, 1 mole of gas occupies 22.4 L. Therefore, the molar volume of the gas is 22.4 L.
Correct Answer: A — 22.4 L
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Q. If 2 moles of an ideal gas at 300 K occupy a volume of 10 L, what is the pressure of the gas? (Use R = 0.0821 L·atm/(K·mol))
A.
0.5 atm
B.
1.0 atm
C.
2.0 atm
D.
3.0 atm
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Solution
Using the Ideal Gas Law, P = nRT/V = (2 moles * 0.0821 L·atm/(K·mol) * 300 K) / 10 L = 4.926 atm.
Correct Answer: B — 1.0 atm
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Q. If 2 moles of glucose are dissolved in 1 liter of water, what is the molarity of the solution?
A.
1 M
B.
2 M
C.
0.5 M
D.
3 M
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Solution
Molarity = moles of solute / liters of solution = 2 moles / 1 L = 2 M.
Correct Answer: B — 2 M
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Q. If 2 moles of KCl are dissolved in 1 liter of solution, what is the molarity of the solution?
A.
1 M
B.
2 M
C.
0.5 M
D.
3 M
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Solution
Molarity (M) = moles of solute / liters of solution = 2 moles / 1 L = 2 M.
Correct Answer: B — 2 M
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Q. If 2 moles of NaCl are dissolved in 1 kg of water, what is the van 't Hoff factor (i)?
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Solution
NaCl dissociates into 2 ions (Na⁺ and Cl⁻), so the van 't Hoff factor (i) is 2.
Correct Answer: C — 3
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Q. If 2 pencils cost 50 cents, how much do 10 pencils cost?
A.
2.50
B.
5.00
C.
7.50
D.
10.00
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Solution
If 2 pencils cost 50 cents, then 10 pencils (5 times more) will cost 5 * 50 = 250 cents or $2.50.
Correct Answer: B — 5.00
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Q. If 20 grams of glucose (C6H12O6) is dissolved in 1 liter of water, what is the molarity of the solution? (Molar mass of glucose = 180 g/mol) (2020) 2020
A.
0.11 M
B.
0.5 M
C.
0.2 M
D.
0.33 M
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Solution
Moles of glucose = 20 g / 180 g/mol = 0.111 M. Molarity = 0.111 moles / 1 L = 0.11 M.
Correct Answer: A — 0.11 M
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Q. If 20 grams of glucose (C6H12O6) is dissolved in 200 mL of solution, what is the mass percent concentration? (Molar mass of glucose = 180 g/mol)
A.
10%
B.
5%
C.
20%
D.
15%
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Solution
Mass percent = (mass of solute / total mass of solution) x 100 = (20 g / (20 g + 200 g)) x 100 = 10%.
Correct Answer: A — 10%
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Q. If 200 g of ice at 0°C is added to 100 g of water at 80°C, what will be the final temperature of the mixture? (Latent heat of fusion of ice = 334 J/g)
A.
0°C
B.
20°C
C.
40°C
D.
60°C
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Solution
Heat lost by water = Heat gained by ice. 100g * 80°C = 200g * 334 J/g + 200g * (Tf - 0°C). Solving gives Tf = 20°C.
Correct Answer: B — 20°C
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