Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

Barium chloride reacts with sodium sulfate to form barium sulfate and sodium chloride. Then, according to the law of conservation of mass:

A
The total mass of reactants, barium chloride, and sodium sulfate taken is greater than the total mass of products, barium sulfate, sodium chloride formed.

B
The total mass of reactants, barium chloride, and sodium sulfate taken is less than the total mass of products, barium sulfate, sodium chloride formed.

C
The total mass of the reactants is equal to the total mass of products.

D
None of the above

Solution

$$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$$
The total mass of the reactants is equal to the total mass of products.
Hence, option C is correct.
Question 2
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A balanced chemical equation supports

A
Law of chemical equilibrium

B
Law of conservation of energy

C
Law of conservation of mass

D
None of these

Solution

A balanced chemical equation supports "Law of conservation of mass". Law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
Question 3
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If 11 g of oxalic acid are dissolved in 500 mL of solution (density = 1.1 g mL$$^{-1}$$), what is the mass % of oxalic acid in solution?

A
1%

B
2%

C
3%

D
4%

Solution

given $$11 g$$ oxalic acid
volume of solution = $$500 ml$$
density $$= 1.1 = \dfrac{mass}{volume}$$
mass of solution =$$ 1.1 \times 500$$
weight % of oxalic acid = $$\dfrac{11}{1.1 \times 500} \times 100$$
$$= 2\%$$
Question 4
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8 g of a solute is dissolved in 92 g of solvent. If the density of solution is 2 g ml$$^{-1}$$, find the weight-volume percentage.

A
10 g ml$$^{-1}$$

B
12 g ml$$^{-1}$$

C
16 g ml$$^{-1}$$

D
15 g ml$$^{-1}$$

Solution

given $$ 8\,g $$ solute
$$ 92\,g\, $$ solvent
solution $$ = 100\,g $$
density $$ = \dfrac{mass}{volume} = 2 $$
$$\dfrac{100}{volume} = 2 $$
volume $$ = 50\,ml $$
weight $$-$$ volume $$\% = \dfrac{weight\,of\,solute}{volume\,of\,sol^{n}}\times 100 = \dfrac{8}{50}\times 100 $$
$$ = 16\,g\,ml^{-1} $$
Question 5
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Solubility of a solute is $$S$$. Find its percentage by weight.

A
$$\dfrac{200S}{S + 100}$$

B
$$\dfrac{50S}{S + 100}$$

C
$$\dfrac{100S}{S + 100}$$

D
None of these

Solution

Solubility of solute is S means that
S grams per 100g solvent.
So weight of solution $$ = (S+100)$$ grams
weight % $$ = \dfrac{weight\,of\,solute}{weight\,of\,solution}\times 100 $$
$$ = \dfrac{100\,S}{5+100} $$
Question 6
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Nitrogen reaction with oxygen to form nitrous oxide.

What will be the volume ratio of the reactants and products of the given reaction?
$$2N_{2}+O_{2}\rightarrow 2N_{2}O$$

A
1 : 1 : 2

B
1 : 2 : 2

C
3 : 1 : 2

D
2 : 1 : 2

Solution

$$2N_{2}+O_{2}\rightarrow 2N_{2}O$$

The Law of chemical combination states that the Ratio of reactants & products are Reactions coefficients or stoichiometry coefficients.
Ratio $$= 2:1:2$$

Hence, the correct option is $$(D)$$.
Question 7
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A solution with a weight percentage of sodium hydroxide of 9% is prepared by adding the following mass of water (in grams) to 300 grams of that solution with a weight percent of $$NaOH$$ of 15%.

A
100 g

B
150 g

C
200 g

D
250 g

Solution

15 = $$\dfrac{NaOH weight}{(total weight)}\times100$$
in this eqution total weight is 300gm so NaOH weight is 45 gm
in this question when x gm of water add in this solution we will get weight percent of NaOH is 9%
so the eqution is
9 =$$\dfrac{45}{300+x}\times100$$
in sove this eqution x is 200gm
so the answer is 200gm
answer is C
Question 8
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How many mL of sulphuric acid of density 1.84 g mL$$^{-1}$$ containing 95.6 mass % of $$H_2SO_4$$ should be added to one litre of 40 mass % solution of $$H_2SO_4$$ of density 1.31 g mL$$^{-1}$$ in order to prepare 50 mass % solution of sulphuric acid of density 1.40 g mL$$^{-1}$$?

A
66.2 ml

B
55.2 ml

C
95.5 ml

D
105.2 ml

Solution

Let mass of solution be 100 gm
Therefore amount of solute in it = 95.6 gm
Volume of solution = $$\dfrac{mass}{density}$$ = $$\dfrac{100}{1.84}$$ = 54.3478 ml

Volume = 1 L =1000 ml
Density = 1.31 g/ml
Mass of solution = 1310 g
Mass % of solute = 40 %

Let mass of solute in it be X g

Mass % = $$\dfrac{mass of solute} {mass of solution} \times100$$

40 = $$\dfrac{X}{1300}\times100$$

X = $${40\times1310}$$ = 524 gm

For solution to make 50 % mass percent and density 1.40 g/ml
Volume of solution =1400 ml
Amount of solute present should be,

Y gm = $$\dfrac{50\times1400}{100}$$ = 700 g

Thus
Amount of solute needed to be added to the 40 % to make it 50% by mass =700-524 = 176 g

Volume of H2SO4 of density 1.84 g/ml required to be added to 40 % = $$\dfrac{mass}{density}$$ = $$\dfrac{176}{1.84}$$ =95.6521 ml
so the answer is C
Question 9
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A student has only 200 g of valuable solvent and wishes to make a solution of 20% by weight from solute $$A$$. How many grams of $$A$$ is weighed out?

A
50 grams

B
250 grams

C
20 grams

D
30 grams

Solution

$$200 g $$ solvent
$$Solute = 20\%$$
$$\dfrac{weight\, of\, solute}{total\, weight}=\dfrac{20}{100}$$
$$\dfrac{x}{200+x}=\dfrac{1}{5}$$
$$x=50$$ grams.
Question 10
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Calculate the number of iron atoms in a piece of iron weighing $$2.8 g$$. (Atomic mass of iron $$=56$$)

A
$$30.11\times { 10 }^{ 23 }$$ atoms

B
$$3.11\times { 10 }^{ 23 }$$ atoms

C
$$3.0115\times { 10 }^{ 22 }$$ atoms

D
$$301.1\times { 10 }^{ 23 }$$ atoms

Solution

Given that, weight of iron =2.8 g
Atomic mass= 56
moles of iron$$= \dfrac{\text{wt. of iron}}{\text{atomic mass}}=\dfrac{2.8}{56}=0.05\ moles$$
1 mole $$= 6.022 \times 10^{23}\ atoms$$
0.05 moles $$= 6.022 \times 0.05 \times 10^{23}\ atoms=3.0115 \times 10^{22}$$ atoms
Option C is correct.

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Some Basic Concepts of Chemistry Test - 38

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Some Basic Concepts of Chemistry Test - 38

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Question 1

The total mass of reactants, barium chloride, and sodium sulfate taken is greater than the total mass of products, barium sulfate, sodium chloride formed.

The total mass of reactants, barium chloride, and sodium sulfate taken is less than the total mass of products, barium sulfate, sodium chloride formed.

The total mass of the reactants is equal to the total mass of products.

None of the above

Solution

Question 2

Law of chemical equilibrium

Law of conservation of energy

Law of conservation of mass

None of these

Solution

Question 3

1%

2%

3%

4%

Solution

Question 4

10 g ml$$^{-1}$$

12 g ml$$^{-1}$$

16 g ml$$^{-1}$$

15 g ml$$^{-1}$$

Solution

Question 5

$$\dfrac{200S}{S + 100}$$

$$\dfrac{50S}{S + 100}$$

$$\dfrac{100S}{S + 100}$$

None of these

Solution

Question 6

1 : 1 : 2

1 : 2 : 2

3 : 1 : 2

2 : 1 : 2

Solution

Question 7

100 g

150 g

200 g

250 g

Solution

Question 8

66.2 ml

55.2 ml

95.5 ml

105.2 ml

Solution

Question 9

50 grams

250 grams

20 grams

30 grams

Solution

Question 10

$$30.11\times { 10 }^{ 23 }$$ atoms

$$3.11\times { 10 }^{ 23 }$$ atoms

$$3.0115\times { 10 }^{ 22 }$$ atoms

$$301.1\times { 10 }^{ 23 }$$ atoms

Solution

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