Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The number of atoms present in 78 g of potassium is equal to the number of potassium ions present in .......... g of potassium chloride.

A
74.5

B
37.25

C
223.5

D
149

Solution

No. of atoms present in 39 g of K = $$N_{A}$$
$$\therefore$$ No. of atoms present in 78 g of K = 2$$N_{A}$$
Now, $$N_{A}$$ ions of potassium are present in 74.5 g of $$KCl $$.
2$$N_{A}$$ ions of potassium are present in 149 g of $$KCl$$.
Question 2
1 / -0

10.0 g of CaCO$$_3$$ on heating gave 4.4 g of CO$$_2$$ and 5.6 g of CaO. The observation is in agreement with the:

A
law of constant composition

B
law of multiple proportions

C
law of reciprocal proportion

D
law of conservation of mass

Solution

$$CaCO_3 \xrightarrow {\triangle} CaO + CO_2$$
$$10.0 \ g$$ $$5.6 \ g$$ $$4.4 \ g$$

Since, mass of reactant = Mass of product

$$10.0 \ g = 4.4 \ g+5.6 \ g$$

$$10.0 \ g= 10.0 \ g$$

So, the law of conservation of mass is constant.
Question 3
1 / -0

What is the percentage by weight of sulphuric acid if 13 g of $$ H_2SO_4$$ is dissolved to make 78 g of solution?

A
23.4

B
20

C
16.6

D
13.2

Solution

Weight of $$H_{2}SO_{4}$$ = 13 g
Weight of solution = 78 g
Therefore (w/W)% = $$\dfrac{13}{78}\times 100$$ $$= 16.6$$%
Question 4
1 / -0

The w/W % of 25 g of calcuim hydroxide in 50 g of solvent is_____ %

A
40

B
33.33

C
36.3

D
30

Solution

Weight of $$Ca(OH)_{2}$$ = 25g

Weight of solution = 75g

Therefore w/W% = $$\dfrac{25}{75}\times 100$$ = 33.33%

Hence, the correct option is $$B$$
Question 5
1 / -0

Which pair has same percentage of carbon?

A
$$CH_3COOH \: and \: C_6H_{12}O_6$$

B
$$CH_3COOH \: and \: C_{12}H_{22}O_{11}$$

C
$$CH_3COOH \: and \: C_2H_5OH$$

D
$$C_6H_{12}O_6 \: and \: C_{12}H_{22}O_{11}$$

Solution

The table below shows the ratio of different atoms present in a molecule.
$$Compound $$ $$C$$ $$H$$ $$O$$
$$CH_3COOH$$ 1 2 1
$$C_6H_{12}O_6$$ 1 2 1
$$C_{12}H_5OH$$ 2 6 1
$$C_{12}H_{22}O_{11}$$ 12; 22; 11
So, $$CH_3COOH$$ and $$C_6H_{12}O_6$$ have same % of $$C$$.
Question 6
1 / -0

The first organic substance ever prepared in the laboratory from the inorganic compound is:

A
urea

B
ethyl alcohol

C
methane

D
ethane

Solution

Urea was the first organic compound to be prepared in the laboratory, which was synthesized by chance. It was prepared by Friedrich Wohler, a German chemist in the year 1828. Wohler synthesized urea from an inorganic compound, ammonium cyanate.
Question 7
1 / -0

What is the mass of the solvent present in 200 g of 25% (w/W) calcium hydroxide solution?

A
150 g

B
125 g

C
175 g

D
100 g

Solution

Given that Mass of solution = 200g and let mass of solute = x
Now, mass of the solute can be calculated as follow: $$25 = \frac{x}{200}\times 100$$
Thus, x = mass of solute = 50g
Therefore mass of solvent = mass of solution - mass of solute = 200-50 = 150g
Question 8
1 / -0

The illustrations show a conservation-of-mass experiment. The solution in the beaker lost mass because

A
materials have less mass at high temperatures

B
the mass of the reactants and products was less than $$100\ g$$

C
sodium sulfate $$(Na_{2}SO_{4})$$ is lighter than air

D
some of the water molecules turned into gas

Solution

The solution in the beaker lost mass as the solution contains water as it is aqueous so due to heat the water has been converted into gas.
The rest three options are not correct explanation.
Question 9
1 / -0

In a chemical reaction, $$100 g$$ baking soda mixture containing sodium bicarbonate and vinegar on heating gives $$43 g$$ of carbon dioxide gas. What mass of solid residue is left in food?

A
$$54\ g$$

B
$$55\ g$$

C
$$56\ g$$

D
$$57\ g$$

Solution

According to law of conservation of mass, total mass of reactants is equal to total mass of products. Here baking soda mixture on heating gives solid residue and carbon dioxide.
$$M_{\text {Baking soda mixture}} = M_{\text {solid residue}} + M_{CO_{2}}$$
or it can be written as:
$$M_{\text {solid residue}} = M_{\text {Baking soda mixture}} - M_{CO_{2}}$$ $$(Eq 1)$$
According to the question
$$M_{\text {Baking soda mixture}} = 100 g$$
$$M_{CO_{2}} = 43 g$$
Substituting the value in equation $$1$$, we get
The mass of solid residue is $$100g - 43g =57 g$$
Question 10
1 / -0

In the procedure shown above, a calcium chloride solution is mixed with a sodium sulfate solution to create the products shown. Which of the following is illustrated by this activity?

A
Law of conservation of mass

B
Law of definite proportions

C
Dalton's law atomic theory

D
Theory of covalent bonding

Solution

In both the diagrams mass of the reactants and products are constant. This goes with the law of conservation of mass.

Hence. option A is correct.

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$$Compound $$	$$C$$	$$H$$	$$O$$
$$CH_3COOH$$	1	2	1
$$C_6H_{12}O_6$$	1	2	1
$$C_{12}H_5OH$$	2	6	1
$$C_{12}H_{22}O_{11}$$	12;	22;	11

Some Basic Concepts of Chemistry Test - 31

Result

Some Basic Concepts of Chemistry Test - 31

Score

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Rank

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SHARING IS CARING

Question 1

74.5

37.25

223.5

149

Solution

Question 2

law of constant composition

law of multiple proportions

law of reciprocal proportion

law of conservation of mass

Solution

Question 3

23.4

20

16.6

13.2

Solution

Question 4

40

33.33

36.3

30

Solution

Question 5

$$CH_3COOH \: and \: C_6H_{12}O_6$$

$$CH_3COOH \: and \: C_{12}H_{22}O_{11}$$

$$CH_3COOH \: and \: C_2H_5OH$$

$$C_6H_{12}O_6 \: and \: C_{12}H_{22}O_{11}$$

Solution

Question 6

urea

ethyl alcohol

methane

ethane

Solution

Question 7

150 g

125 g

175 g

100 g

Solution

Question 8

materials have less mass at high temperatures

the mass of the reactants and products was less than $$100\ g$$

sodium sulfate $$(Na_{2}SO_{4})$$ is lighter than air

some of the water molecules turned into gas

Solution

Question 9

$$54\ g$$

$$55\ g$$

$$56\ g$$

$$57\ g$$

Solution

Question 10

Law of conservation of mass

Law of definite proportions

Dalton's law atomic theory

Theory of covalent bonding

Solution

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