Some Basic Concepts Of Chemistry Test 16

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Some Basic Concepts Of Chemistry Test 16

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Weekly Quiz Competition

Question 1
1 / -0

1 g of Mg is burnt in a closed vessel containing 0.5 g of O₂. Which reactant is limiting reagent and how much of the excess reactant will be left?

A
O₂ is a limiting reagent and Mg is in excess by 0.25 g

B
Mg is a limiting reagent and is in excess by 0.5 g

C
O₂ is a limiting reagent and is in excess by 0.25 g

D
O₂ is a limiting reagent and Mg is in excess by 0.75 g

Solution

48 g of Mg requires 32 g of O₂
1 g of Mg requires 32/48 = 0.66 g of O₂
Oxygen available = 0.5 g
Hence, O₂ is limiting reagent.
32 g of O₂ reacts with 48 g of Mg
0.5 g of O₂ will react with 48/32 x 0.5 = 0.75 g of Mg
Excess of Mg = (1.0 - 0.75) = 0.25 g
Question 2
1 / -0

In a reaction container, 100 g of hydrogen and 100 g of CI₂ are mixed for the formation of HCl gas. What is the limiting reagent and how much HCl is formed in the reaction?

A
H₂ is limiting reagent and 36.5 g of HCl are formed

B
CI₂ is limiting reagent and 102.8 g of HCl are formed

C
H₂ is limiting reagent and 142 g of HCl are formed

D
CI₂ is limiting reagent and 73 g HCl are formed

Solution

2 g of H₂reacts with 71 g of CI₂
100 g of H₂ will react with 71/2 x 100 = 3550g of CI₂
Hence, CI₂ is the limiting reagent.
71 g of CI₂ produces 73 g of HCl
100g of CI₂ will produce 73/71 x 100 = 102.8 g of HCl
Question 3
1 / -0

If 40 g of CaCO₃is treated with 40 g of HCl, which of the reactants will act as limiting reagent?

A
CaCO₃

B
HCl

C
Both are equal

D
Cannot be calculated

Solution

100 g of CaCO₃ reacts with 73 g of HCl
40 g of CaCO₃ will react with 73/100 x 40 = 29.2 g of HCl
Since CaCO₃ is completely consumed and some amount (40 - 29.2 = 10.8g) of HCl remains unreacted and hence, CaCO₃ is limiting reagent.
Question 4
1 / -0

The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO₃ is

A
28 g

B
9.25 g

C
2.870 g

D
58 g

Solution

AgNO₃ + NaCl → AgCl + NaNO₃
No. of moles of AgNO₃ = 3.4/170 = 0.02
No, of moles of NaCl = 5.85/58.5 = 0.1
Limiting reagent = AgNO₃
1 mole of AgNO₃ produces 1 mole of AgCl
0.02 mole of AgNO₃ will produce 0.02 mole of AgCl
Weight of AgCl produced = 0.02 x 143.5 = 2.870 g.
Question 5
1 / -0

How much oxygen is required for complete combustion of 560 g of ethene?

A
6.4 kg

B
1.92 kg

C
2.8 kg

D
9.6 kg

Solution

28 g of C₂H₄ requires 96 g of O₂
560 g of C₂H₄ requires 96/28 x 560
= 1920 g or 1.92 kg of O₂
Question 6
1 / -0

What is the mass percent of oxygen in ethanol?

A
52.14%

B
13.13%

C
16%

D
34.73%

Solution

Molecular formula of ethanol = C₂H₅OH
Molar mass of ethanol of 2 x 12.01 + 6 x 1.008 + 16 = 46.068g
Mass percent of oxygen = 16/46.068 x 100 = 34.73%
Question 7
1 / -0

How much mass of sodium acetate is required to make 250 mL of 0.575 molar aqueous solution?

A
11.79 g

B
15.38 g

C
10.81 g

D
25.35 g

Solution

Molar mass of sodium acetate (CH₃COONa)
= 82.0245 g/mol
Mass of CH₃COONa required to make 250 mL of 0.575 M solution
Question 8
1 / -0

How much copper is present in 50 g of CuSO₄?

A
19.90 g

B
39.81 g

C
63.5 g

D
31.71 g

Solution

Molar mass of CuSO₄ = 63.5 + 32 + 4 x 16 = 159.5 g
Mass of copper present in 159.5 g of CuSO₄ = 63.5 g
∴ Mass of copper present in 50 g of CuSO₄
Question 9
1 / -0

A solution is prepared by adding 5 g of a solute 'X' to 45 g of solvent 'Y'. What is the mass per cent of the solute 'X'?

A
10%

B
11.1%

C
90%

D
75%

Solution

Mass percent of X
Question 10
1 / -0

An impure sample of silver (1.5 g) is heated with S to form 0.124 g of Ag₂S. What was the per cent yield of Ag₂S?

A
21.6%

B
7.2%

C
1.7%

D
24.8%

Solution

216 g of Ag forms 248 g of Ag₂S
1.5 g of Ag forms 248/216 x 1.5 = 1.722 g of Ag₂S
% yield of Ag₂S = 0.124/1.722 x 100 = 7.2%
Question 11
1 / -0

2.82 g of glucose is dissolved in 30 g of water. The mole fraction of glucose in the solution is

A
0.01

B
0.99

C
0.52

D
1.66

Solution

No. of moles of glucose = 2.82/180 = 0.01567
No. of moles of water = 30/18 = 1.667
Total no. of moles of solution = 0.01567 + 1.667 = 1.683
Mole fraction of glucose = 0.01567/1.683 = 0.0093 = 0.01
Question 12
1 / -0

What volume of water is to be added to 100 cm³ of 0.5 M NaOH solution to make it 0.1 M solution?

A
200 cm³

B
400 cm³

C
500 cm³

D
100 cm³

Solution

M₁V₁= M₂V₂
0.5 x 100 = 0.1 x V₂
V₂ = 500 cm³
Volume of water to be added to 100 cm³ of solution
= 500 - 100 = 400 cm³
Question 13
1 / -0

The final molarity of a solution made by mixing 50 mL of 0.5 M HCl, 150 mL of 0.25 M HCl and water to make the volume 250 mL is

A
0.5 M

B
1 M

C
0.75 M

D
0.25 M

Solution

M₁V₁= M₂V₂= MV
0.5 x 50 + 0.25 x 150 = M x 250
M = (25 + 37.5)/250 = 0.25 M
Question 14
1 / -0

A solution is made by dissolving 49 g of H₂SO₄ in 250 mL of water. The molarity of the solution prepared is

A
2 M

B
1 M

C
4 M

D
5 M

Solution

Molarity

= 2 M
Question 15
1 / -0

What is the concentration of copper sulphate (in mol L^-1) if 80 g of it is dissolved in enough water to make a final volume of 3 L?

A
0.0167

B
0.167

C
1.067

D
10.67

Solution

Molar mass of CUSO₄ = 63.5 + 32 + 64 = 159.5
Moles of CUSO₄ = 80/159.5 = O.50
Volume of solution = 3 L

= 0.167 mol L^-1
Question 16
1 / -0

4.28 g of NaOH is dissolved inwater and the solution is made to 250 cc. What will be the molarity of the solution?

A
0.615 mol L^-1

B
0.428 mol L^-1

C
0.99 mol L^-1

D
0.301 mol L^-1

Solution

No. of moles of NaOH = 4.28/40 = 0.107 Volume of solution = 250 cm³
M = n/V(in L) = 0.107/250 x 1000 = 0.428 mol L^-1
Question 17
1 / -0

What volume of 5 M Na₂SO₄ must be added to 25 mL of 1 M BaCl₂ to produce 10 g of BaSO₄?

A
8.58 mL

B
7.2 mL

C
10 mL

D
12 mL

Solution

Na₂SO₄ + BaCI₂ → BaSO₄ + 2NaCl
No. of moles of BaSO₄ = w/M = 10/233 = 0.0429
∴ No. of moles of Na₂SO₄ needed = M x V/1000
Or 0.0429 = 5 x V/1000
V = 8.58 mL
Question 18
1 / -0

What will be the molarity of the solution in which 0.365 g of HCl gas is dissolved in 100 mL of solution?

A
2 M

B
0.2 M

C
1 M

D
0.1 M

Solution

No. of moles in 0.365 g of HCl = 0.365/36.5 = 0.01
Volume of solution in L = 100/1000 = 0.1 L
Molarity = n/V (in L) = 0.01/0.1 = 0.1 M
Question 19
1 / -0

What will be the molality of the solution made by dissolving 10 g of NaOH in 100 g of water?

A
2.5 m

B
5 m

C
10 m

D
1.25 m

Solution

Molality =
Question 20
1 / -0

What will be the molality of chloroform in the water sample which contains 15 ppm chloroform by mass?

A
1.25 x 10^-4m

B
2.5 x 10^-4 m

C
1.5 x 10^-3 m

D
1.25 X 10^-5 m

Solution

15 ppm = 15/10⁶ x 10² = 1.5 x 10^-3g
Molality of CHCl₃ solution = 1.5 x 10^-3/100 x 1000/119.5
= 1.25 x 10^-4m