Some Basic Concepts Of Chemistry Test 7

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Some Basic Concepts Of Chemistry Test 7

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Weekly Quiz Competition

Question 1
1 / -0

For the reaction 2x + 3y + 4z → 5w
Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is

A
50%

B
60%

C
70%

D
40%

Solution

1 moles of x will give = 5/2 = 2.5 mol
Question 2
1 / -0

Molarity of NaOH in a solution prepared by dissolving 4 g of NaOH in enough water to form 250 ml of solution is:

A
1.4 M

B
0.45 M

C
0.4 M

D
4.1 M

Solution
Question 3
1 / -0

125 ml of 8% w/w NaOH solution (sp. gravity 1) is added to 125 ml of 10% w/v HCl solution. The nature of resultant solution would be ____

A
Acidic

B
Basic

C
Neutral

D
None

Solution

= 0.25 mole
Question 4
1 / -0

Ratio of masses of H₂SO₄ and Al₂ (SO₄)₃ is grams each containing 32 grams of S is _____

A
0.86

B
1.72

C
0.43

D
2.15

Solution
Question 5
1 / -0

18 g of glucose (C₆H₁₂O₆) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:

A
1M

B
0.1M

C
1.1M

D
0.5M

Solution

18 g of glucose (C₆H₁₂O₆) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is 0.1 M.
Question 6
1 / -0

For the reaction 2A + 3B + 5C → 3D
Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken, With 25% yield, moles of D which can be produced are _____________.

A
0.75

B
0.5

C
0.25

D
0.6

Solution

Limiting reactant is A
Ideally with 2 moles of A, D formed = 3 moles But yield = 25%
So, moles of D formed = 3 × 0.25 = 0.75 mol
Question 7
1 / -0

Two elements X (atomic mass = 75) and Y (atomic mass = 16) combine to give a compound having 75.8% of X. The formula of the compound is:

A
X₂Y₃

B
X₂Y

C
X₂Y₂

D
XY

Solution

So empirical formula of compound = X₂Y₃
For a molecular formula you have to provide molecular mass of the compound.
Question 8
1 / -0

Equal volumes of 10% (v/v) of HCl is mixed with 10% (v/v) NaOH solution. If density of pure NaOH is 1.5 times that of pure HCl then the resultant solution be :

A
Basic

B
Neutral

C
Acidic

D
Can’t be predicted.

Solution

Both have equal volume = V

NaOH mole > HCl mole
Basic Solution
Question 9
1 / -0

Similar to the % labelling of oleum, a mixture of H₃PO₄ and P₄O₁₀ is labelled as (100 + x) % where x is the maximum mass of water which can react with P₄O₁₀ present in 100 gm mixture of H₃PO₄ and P₄O₁₀. If such a mixture is labelled as 127% Mass of P₄O₁₀ is 100 gm of mixture, is

A
71 gm

B
47 gm

C
83 gm

D
35 gm

Solution

P₄O₁₀ + 6 H₂O → 4H₃ PO₄
284 gm 108gm 392 gm 108 gm water reacts with P₄O₁₀ = 284 gm
27 gm water will react with P₄O₁₀ = = 71 gm
Question 10
1 / -0

C₆H₅OH(g) + O₂(g) → CO₂(g) + H₂O(l)
The magnitude of volume change if 30 ml of C₆H₅OH (g) is burnt with an excess amount of oxygen, is

A
30 ml

B
60 ml

C
20 ml

D
10 ml

Solution

C₆H₅OH(g)+7O₂(g) → 6CO₂(g)+3H₂O(l)
30 ml
6 x 30= 180 ml of CO₂ is produced
Volume used initially = 30 + 210 = 240
(for C₆H₅OH)(For O₂)
change in volume = 240 - 180 = 60 ml
Question 11
1 / -0

How many moles of NaCl should be dissolved in 100 mL of water to get 0.2 M solution?

A
0.2

B
0.02

C
0.01

D
0.1

Solution

Molarity = no of moles /volume( in litre) 0.2 =no of moles / 0.1 No of moles =0.2 × 0.1 = 0.02
Question 12
1 / -0

If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option is

A
The resulting solution contains 18 gm of water and 118 gm H₂SO₄

B
The resulting solution contains 9 gm of water and 59 gm H₂SO₄

C
The resulting solution contains only 118 gm pure H₂SO₄

D
The resulting solution contains 68 gm of pure H₂SO₄

Solution

118 % ⇒ 100 g sample uses 18 g water
⇒ 50 g sample need 9 g water
(50 g + 18 g water)
⇒ 59 g H₂SO₄ + 9 g water
Question 13
1 / -0

The minimum mass of mixture of A₂ and B₄ required to produce at least 1 kg of each product is:
(Given At. mass of ‘A’ = 10; At mass of ‘B’ = 120) 5A₂ + 2B₄ → 2AB₂ + 4A₂B

A
2120 gm

B
1060 gm

C
560 gm

D
1660 gm

Solution

Here limiting product is AB2 = 500 gm

So, A₂ needed = 10 × 20 = 200 gm
B₂ needed = 480 × 4 = 1920 gm
Total mass of mixture = 2120 gm
Question 14
1 / -0

The % by volume of C₄H₁₀ in a gaseous mixture of C₄H₁₀, CH₄ and CO is 40. When 200 ml of the mixture is burnt in excess of O₂. Find volume (in ml) of CO₂ produced.

A
220

B
340

C
440

D
560

Solution

C₄H₁₀ = 80 ml
CH₄ = xml CO = y ml
x + y = 120 ml

total CO₂ volume
= 320 + x + y ml
= 320 + 120
= 440 ml
Question 15
1 / -0

74 gm of sample on complete combustion gives 132 gm CO₂ and 54 gm of H₂O. The molecular formula of the compound may be

A
C₅H₁₂

B
C₄H₁₀O

C
C₃H₆O₂

D
C₃H₇O₂

Solution

⇒ C atoms = 3 mole
H atoms = 6 mole
Question 16
1 / -0

Density of a gas relative to air is 1.17. Find the mol. mass of the gas. [M_air = 29 g/mol]

A
33.9

B
24.7

C
29

D
22.3

Solution

M gas = 29 × 1.17 = 33.9
Question 17
1 / -0

Number of moles of hydroxide ions in 0.3L of 0.005M solution of Ba (OH)₂ is:

A
0.0015

B
0.003

C
0.0050

D
0.0075

Solution

0.3 L = 0.3 L of 0.005 M Ba(OH)₂
0.3 x 0.005 = 0.0015 moles of Ba(OH)₂
It will dissociate into 0.0015 moles of Ba and 2 x 0.0015 = 0.003 moles of OH ions.