Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

When 100 ml of a $$O_{2} - O_{3}$$ mixture was passed through turpentine, there was the reduction of volume by 20 ml. If 100ml of such a mixture is heated, what will be the increase in volume?

A
$$10\ ml$$

B
$$20\ ml$$

C
$$30\ ml$$

D
$$25\ ml$$

Solution

Ozone is highly soluble in turpentine. Thus, the decrease in volume of $$20$$ ml is due to the solubility of ozone. Thus $$100 ml$$ of a mixture contains $$20$$ ml of ozone.

On heating, $$20$$ ml of ozone decomposes to form $$30$$ ml of oxygen. The net increase in volume will be $$(30 - 20)=10 ml$$.

$$2O_3 \rightarrow 3O_2$$

So, the correct option is $$A$$
Question 2
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The mole fraction of solute in aqueous urea solution is $$0.2$$. The mass percent of solute is:

A
$$45.45\%$$

B
$$43.34\%$$

C
$$34.23\%$$

D
$$35.67\%$$

Solution

The mole fraction of urea is $$0.2$$.

The mole fraction of water is $$1-0.2=0.8$$.

The molar masses of urea and water are $$60$$ g/mol and $$18$$ g/mol respectively.

Let us say one mole of the total solution is present.

The number of moles of urea and water present will be $$0.2$$ mol and $$0.8$$ mol. Their masses will be $$0.2 \times 60 = 12$$ g urea and $$0.8 \times 18=14.4$$ g water.

The mass percentage of solution $$= \dfrac {Mass\ of \ urea} {Total \ mass\ of\ solution} = \dfrac {12} {12+14.4} \times 100=45.45$$ %.
Question 3
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The vapours of organic compound was burnt in oxygen. Equal volume of both gaseous substance were taken at same pressure and temperature. After the reaction, the system was returned to the original condition and it turn out that its volume has not changed. The product of combustion contain 50% $$CO_{2}$$(g) and 50% $$H_{2}O$$(g) by volume and no other gas. Find the molecular weight of organic compound (in gram/mol) in question.

A
30

B
35

C
40

D
46

Solution

Let the organic compound be X.

The volume of a gas is directly proportional to the number of moles.

According to the given data, 1 mole of X combines with 1 mole of oxygen to form 1 mole of carbon dioxide and 1 mole of water.

$$X+O_2 \rightarrow CO_2 + H_2O$$

Thus the molecular formula of X is $$CH_2O$$.

Its molecular weight is $$12+2+16=30$$ g/mol

Hence, the correct option is $$\text{A}$$
Question 4
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10 ml of CO is mixed with 25 ml air having 20 per cent $$O_{2}$$ by volume. What would be the final volume if none of CO and $$\displaystyle\: O_{2}$$ is left after the reaction?

A
30 ml

B
40 ml

C
43 ml

D
49 ml

Solution

The air contains approximately 20% oxygen and 80% nitrogen.

Thus 25 ml of air contains 5 ml oxygen and 20 ml nitrogen.

The reaction for the conversion of CO is as shown below.

$$\underset {10 ml} {CO} + \underset {5 ml} {\cfrac {1} {2}} O_2 \rightarrow \underset {10 ml} {CO_2}$$

Hence, the volume of the mixture after reaction is $$10\ ml\ CO_2 + 20\ ml\ N_2 =30\ ml.$$

Hence, the correct answer is option $$\text{A}$$.
Question 5
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Directions For Questions

The strength of $$H_{2}O_{2}$$ is expressed in several ways like molarity, normality, % (w/v), volume strength etc. The strength of $$10\:V$$ means $$1$$ volume of $$H_{2}O_{2}$$ on decomposition gives $$10$$ volumes of oxygen at $$STP$$ or $$1$$ litre of $$H_{2}O_{2}$$ gives $$10$$ litre of $$O_{2}$$ at $$STP$$. The decomposition of $$H_{2}O_{2}$$ is shown as under:

$$H_{2}O_{2} (aq) \rightarrow H_{2}O(l)+\displaystyle \frac {1}{2} O_{2} (g)$$

$$H_{2}O_{2}$$ can acts as oxidising as well as reducing agent. As oxidizing agent, $$H_{2}O_{2}$$ converted into $$H_{2}O$$ and as reducing agent, $$H_{2}O_{2}$$ converted into $$O_{2}$$. In both the cases, its $$n$$-factor is $$ 2$$.
$$\therefore$$ Normality of $$H_{2}O_{2}$$ solution $$ = 2 \times $$Molarity of $$H_{2}O_{2}$$ solution.

...view full instructions

$$40\: g\: Ba(MnO_{4} )_{2}$$ (mol. wt. $$= 375$$ g/mol) sample containing some inert impurities in acidic medium is completely reacted with $$125$$ mL of $$33.6\: V$$ of $$H_{2}O_{2}$$. What is the percentage purity of the sample?

A
$$28.12 \%$$

B
$$70.31 \%$$

C
$$85 \%$$

D
None of the above

Solution

Mass of $$H_2O_2$$ in $$125$$ ml solution $$=\dfrac{17}{5600}\times 33.6 \times 125$$ g
So, number of equivalents of $$H_2O_2 = 33.6 \times \dfrac{125}{5600} $$
As number of equivalents of both are same,
$$\dfrac{x}{\dfrac{375}{10}} = 33.6\times \dfrac{125}{5600}$$
or, $$ x= 28.125 $$
So, percentage purity of the sample $$= \dfrac{28.125}{40}\times 100 = 70.31\%$$
Question 6
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Calculate the volume of $$CO_{2}$$ evolved by the combustion of $$50$$ ml of a mixture containing $$40$$ per $$C_{2}H_{4}$$ and $$60$$ per $$CH_{4}$$ (by volume).

A
$$70$$ ml

B
$$75$$ ml

C
$$80$$ ml

D
$$82$$ ml

Solution

The volume of $$C_2H_4$$ in $$50$$ ml of sample is $$50 ml \times \dfrac {40} {100} =20 ml$$.

The volume of $$CH_4$$ in $$50$$ ml of sample is $$50 ml \times \dfrac {60} {100} =30 ml$$.

$$\underset {30 ml}{CH_4} \rightarrow \underset {30 ml}{CO_2} + \underset {60 ml} {2H_2O} $$

$$\underset {20 ml}{C_2H_4} \rightarrow \underset {40}{2CO_2} + \underset {40 ml} {2H_2O} $$
The volume of carbon dioxide formed is $$30 ml + 40 ml=70 ml$$.
Question 7
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10 ml of a mixture of $$CH_{4}, C_{2}H_{4}$$ and $$CO_{2}$$ were exploded with excess of air. After explosion, there was contraction on cooling of 17 ml and after treatment with KOH, there was further reduction of 14 ml. What is the composition of the mixture?

A
$$CH_{4}= 4.5 ml, CO_{2}= 1.5 ml$$

B
$$CH_{4}= 3.6 ml, CO_{2}= 2.5 ml$$

C
$$CH_{4}= 4.8 ml, CO_{2}= 1.8 ml$$

D
$$CH_{4}= 5.1 ml, CO_{2}= 2.8 ml$$

Solution

The contraction on cooling is due to the condensation of water vapour. Thus 17 ml of water are formed in the reaction.

The contraction in the volume of 14 ml is due to the dissolution of carbon dioxide gas. Hence, 14 g of carbon dioxide is present in the mixture after the explosion.

$$\underset {4.5 ml}{CH_4} \rightarrow \underset {4.5 ml}{CO_2} + \underset {9 ml} {2H_2O} $$

$$\underset {4 ml}{C_2H_4} \rightarrow \underset {8}{2CO_2} + \underset {8 ml} {2H_2O} $$

Hence, the composition of the mixture is 4.5 ml methane and 1.5 ml of carbon dioxide.
Question 8
1 / -0

A mixture of $$NH_{4}NO_{3}$$ and $$(NH_{4})_{2}HPO_{4}$$ contains $$30.40 \%$$ mass percent of nitrogen. What is the mass ratio of the two components in the mixture?

A
$$2:1$$

B
$$1:2$$

C
$$3:4$$

D
$$4:1$$

Solution

Let weight of $$NH_{4}NO_{3}$$ and $$(NH_{4})_{2}HPO_{4}$$ are $$x$$ and $$y$$ g respectively.
So, mass per cent of nitrogen is given by,
$$\displaystyle \frac{\displaystyle \frac{x}{80} \times 2 \times 14+\displaystyle \frac{y}{132} \times 2 \times 14}{x+y} \times 100=30.4$$

$$\Rightarrow x:y=2:1$$
Question 9
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40 ml of mixture of $$C_{2}H_{2}$$ and CO is mixed with 100 ml of $$O_{2}$$ gas and the mixture is exploded. The residual gases occupied 104 ml and when these are passed through KOH solution, the volume becomes 48ml. All the volume are at same the temperature and pressure. Determine the composition of the original mixture.

A
$$\displaystyle\:C_{2}H_{2}= 16 ml, CO= 24 ml$$

B
$$\displaystyle\:C_{2}H_{2}= 23 ml, CO= 14 ml$$

C
$$\displaystyle\:C_{2}H_{2}= 30 ml, CO= 26 ml$$

D
$$\displaystyle\:C_{2}H_{2}= 33 ml, CO= 28 ml$$

Solution

Let the original mixture contains x ml of CO and (40-x) ml of $$C_2H_2$$.
Th balanced chemical equations are as shown below.
$$\underset {40-x}{C_2H_2} + \underset {2.5 (40-x)} {\dfrac {5}{2}O_2} \rightarrow \underset {2(40-x)} {2CO_2}+H_2O(l)$$
$$\displaystyle\underset {x} CO+ \underset {0.5x}{\dfrac {1} {2}O_2} \rightarrow \underset {x}{CO_2} $$
Total volume is 104 ml.
Hence, $$2(40-x)+x+(100-2.5(40-x)-0.5x=104$$.
On solving this, we get $$x=24\ ml$$.
Hence, the volumes of CO and $$C_2H_2$$ are 24 ml and $$40-24=16$$ ml respectively.
Question 10
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A $$0.60$$ g nitrogen containing compound was boiled with $$NaOH$$, and $$NH_3$$ thus formed, required $$100$$ mL of $$0.2$$ N $$H_2SO_4$$ for neutralisation. The percentage of nitrogen in the compound is:

A
$$46.67\%$$

B
$$23.34\%$$

C
$$60.00\%$$

D
$$20.00\%$$

Solution

100 mL of 0.2 N $$H_2SO_4 \equiv $$ 0.02 equivalents of ammonia.
Since, basicity of ammonia is 1, moles of ammonia is also equal to 0.02 mol.
Each mole of ammonia has 1 mole of nitrogen atom, so moles of nitrogen atom is also equal to 0.02 .

So, 0.60 g of organic compound contains $$\displaystyle \frac {14 \times 20}{1000} $$ g of nitrogen.

Therefore, % of nitrogen in the compound $$ = \dfrac{Mass\ of\ nitrogen}{Mass\ of \ compound}\times 100=\displaystyle \frac {14 \times 20 \times 100}{1000 \times 0.6} = 46.67 $$ % of nitrogen.

So, percentage of nitrogen is $$46.67$$ %

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Some Basic Concepts of Chemistry Test - 66

Result

Some Basic Concepts of Chemistry Test - 66

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SHARING IS CARING

Question 1

$$10\ ml$$

$$20\ ml$$

$$30\ ml$$

$$25\ ml$$

Solution

Question 2

$$45.45\%$$

$$43.34\%$$

$$34.23\%$$

$$35.67\%$$

Solution

Question 3

30

35

40

46

Solution

Question 4

30 ml

40 ml

43 ml

49 ml

Solution

Question 5

$$28.12 \%$$

$$70.31 \%$$

$$85 \%$$

None of the above

Solution

Question 6

$$70$$ ml

$$75$$ ml

$$80$$ ml

$$82$$ ml

Solution

Question 7

$$CH_{4}= 4.5 ml, CO_{2}= 1.5 ml$$

$$CH_{4}= 3.6 ml, CO_{2}= 2.5 ml$$

$$CH_{4}= 4.8 ml, CO_{2}= 1.8 ml$$

$$CH_{4}= 5.1 ml, CO_{2}= 2.8 ml$$

Solution

Question 8

$$2:1$$

$$1:2$$

$$3:4$$

$$4:1$$

Solution

Question 9

$$\displaystyle\:C_{2}H_{2}= 16 ml, CO= 24 ml$$

$$\displaystyle\:C_{2}H_{2}= 23 ml, CO= 14 ml$$

$$\displaystyle\:C_{2}H_{2}= 30 ml, CO= 26 ml$$

$$\displaystyle\:C_{2}H_{2}= 33 ml, CO= 28 ml$$

Solution

Question 10

$$46.67\%$$

$$23.34\%$$

$$60.00\%$$

$$20.00\%$$

Solution

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