Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 72

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Question 1
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0.5g sample of a sulphite salt was dissolved in 20ml solution and 20ml of this solution required 10ml of 0.002M acidified permanganate solution. Hence, the $$\%$$ by mass of sulphite in the sulphite salt is:

A
$$2\%$$

B
$$4\%$$

C
$$6\%$$

D
$$8\%$$
Question 2
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The percentage of $$Se$$ in peroxidase anhydrous enzyme is $$0.5$$% by weight (atomic weight $$= 78.4$$). Then minimum molecular weight of peroxidase anhydrous enzyme is:

A
$$1.568\times 10^{4}$$

B
$$1.568\times 10^{3}$$

C
$$15.68$$

D
$$3.136\times 10^{4}$$
Solution
Correct Option: A

Hint: $$Concentration_{ByMass}$$ $$Percentage = \dfrac{Mass_{solute}}{Mass_{solvent}}\times 100$$

Explanation

Step 1: Preparing data
- Let mass of $$Peroxidase$$ $$anhydrous$$ $$enzyme = x$$ $$gram$$ ...(1)
- Mass percentage of $$Se = 0.5 \%$$ ...(2)
- Atomic mass of $$Se = 78.4g$$ ...(3)
- Therefore, the minimum amount of an element which can be present in a substance is $$1$$ atom ...(4)
- Thus, minimum mass that can be present is atomic mass.
- Therefore, minimum molecular weight of peroxidase anhydrous enzyme is that which contains mass of $$Se = Atomic$$ $$Mass$$
Step 2: Solving using data
- As $$Concentration_{ByMass}$$ $$Percentage = \dfrac{Mass_{solute}}{Mass_{solvent}}\times 100$$ ...(5)
- Putting value from (1), (2), (3) and (4) in (5), we have
- $$0.5 = \dfrac{78.4}{x}\times 100$$
- $$x = \dfrac{784}{5} \times 100 = 1.568 \times 10^4$$
Hence, minimum molecular mass of $$Se$$ present in $$peroxidase$$ $$anhydrous$$ $$enzyme$$ is $$1.568 \times 10^4$$
Question 3
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The amonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M $${ H }_{ 2 }S{ O }_{ 4 }$$ solution. The weight percentage of nitrogen in the compound is

A
$$28$$

B
$$14$$

C
$$56$$

D
$$7$$

Solution

Option A is correct
Question 4
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During the preparation of $$H_{2}S_{2}O_{8}$$ (per disulphuric acid) $$O_{2}$$ gas also releases at anode as by product. When $$9.72\ L$$ of $$H_{2}$$ releases at cathode and $$2.35\ L\ O_{2}$$ at anode at STP, the weight of $$H_{2}S_{2}O_{8}$$ produced in gram is:

A
$$87.12$$

B
$$43.56$$

C
$$83.42$$

D
$$51.74$$

Solution

Option B is correct
Question 5
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The mass of a mixture containing HCl and $$H_2SO_4$$ is 0.1 g. On treatment with an excess of an $$AgNO_3$$ solution, reacted with this acid mixture gives 0.1435 g of AgCl. Mass % of the $$H_2SO_4$$ mixture is :

A
30

B
60

C
59

D
None of these

Solution

Option D is correct i.e none of these because the answer is 63.5
Question 6
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If $$224ml$$ of triatomic gas has a mass of $$1g$$ at $$273K$$ and $$1atm$$ pressure, then the mass of one atom is-

A
$$8.30\times {10}^{-23}gm$$

B
$$2.08\times {10}^{-23}gm$$

C
$$5.53\times {10}^{-23}gm$$

D
$$6.24\times {10}^{-23}gm$$

Solution

$$1$$ mole gas occupies $$22.4$$ litres $$= 22,400$$ ml at STP condition.

$$224$$ ml will then be occupied by $$0.01$$ mol

The weight of 0.01 mol $$= 1$$ g

1 mole of the given gas weighs $$= 100$$ g

$$ 6.023 \times 10^{23} \times 3 $$ (since its trimolecular) atoms has weight of 100 g.

Mass of one atom $$ = \dfrac {100g}{6.023 \times 10^{23} \times 3} $$

$$= 0.0553 \times 100g = 5.53 \times 10^{-23}g$$

Hence, option C is correct.
Question 7
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5 g of a sample of bleaching powder is treated with excess acetic acid and KI solution. The liberated $$I_2$$ required 50 ml of N/10 hypo solution. The % available chlorine in sample is

A
3.55

B
7.0

C
35.5

D
28.2

Solution
Question 8
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When a mixture of $$NaBr$$ and $$NaCl$$ is repeatedly digested with sulphuric acid, all the halogens are expelled and $$Na_{2}SO_{4}$$ is formed quantitatively. With a particular mixture, it was found that the weight of $$Na_{2}SO_{4}$$ obtained was precisely the same as the weight of $$NaBr - NaCl$$ mixture taken. Calculate the ratio of the weights of $$NaCl$$ and $$NaBr$$ in the mixture.

A
$$1:1.454 $$

B
$$2:1.31$$

C
$$1.454 : 1$$.

D
$$1.31:2$$

Solution

$$ \begin{array}{l} \text { NaBr + } \quad \mathrm{H}_{2} \mathrm{SO}_{4} \quad \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \\ \text { a } \mathrm{g} \\ \mathrm{NaCl +} \quad \mathrm{H}_{2} \mathrm{SO}_{4} \quad \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \\ \text { b } \mathrm{g} \end{array} $$
$$ \begin{aligned} \text { Meq of } \mathrm{NaBr}+\text { meq of } \mathrm{NaCl} &=\text { meg of } \mathrm{Na}_{2} \mathrm{SO}_{4} \\ \text { meq of } \mathrm{Na}_{2} \mathrm{SO}_{4} \text { formed }=&(9 / 103) \times 1000+(\mathrm{b} / \mathrm{} 8.5) \times 1000 \\ =& 9.71 \mathrm{a}+17.09 \mathrm{~b} \\ & \Rightarrow \frac{\omega}{142 / 2} \times 1000=9.71 \mathrm{a}+17.09 \mathrm{~b} \\ & \Rightarrow \quad \omega=\frac{142}{2000}(9.71 \mathrm{a}+17.09 \mathrm{~b}) \end{aligned} $$
$$ \begin{array}{l} \text { Also given, } \\ \text { mass of } \mathrm{NaCl}+\text { mass of } \mathrm{NaBr}=\text { mass of } \mathrm{Na}_{2} \mathrm{SO}_{4} \text { formed. } \\ \qquad \begin{aligned} a+b &=\frac{142}{2000}(9.7 a+17.09 \mathrm{~b}) \\ \frac{b}{a}=\frac{1.455}{1} &=1.455: 1 \end{aligned} \\ (\mathrm{C}) \text { is correct option } \end{array} $$
Question 9
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What is the mass percentage of carbon tetrachloride if $$22g$$ of benzene is dissolved in $$122g$$ of carbon tetrachloride?

A
$$84.72$$%

B
$$15.28$$%

C
$$50$$%

D
$$44$$%

Solution

Given:-
Mass of $$C_6H_6 = 22g$$

Mass of carbon tetrachloride $$CCl_4 = 122g$$

Formula:-

Mass % of benzene $$=\dfrac{Mass \ of \ solute }{Total\ mass\ of\ solution }\times 100$$

Mass of solute $$=\dfrac{22}{22 + 122}\times 100= 15.28\% $$

Mass percentage of $$CCl_4 = 100 -$$ mass % of $$C_6H_6$$

$$= 100 - 15.28 =84.72\% $$
Question 10
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Which one of the following is not required for the formation of photochemical smog?

A
Nitrogen oxide

B
Methane

C
Volatile organic compounds

D
Ozone

Solution