Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 21

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Question 1
1 / -0

$$10.0\ g$$ of $$CaCO_{3}$$ on heating gave $$4.4\ g$$ of $$CO_{2}$$ and $$5.6\ g$$ of $$CaO$$. This observation is in agreement with the:

A
law of constant composition

B
law of multiple proportion

C
law of reciprocal proportion

D
law of conservation of mass

Solution

$$CaCO_{3}\rightarrow CaO + CO_{2}$$
10 gm 5.6 gm 4.4 gm

Mass of $$CaCO_{3}\: (10\ gm) =$$ Mass of $$CaO\ (5.6\ gm)$$ + Mass of $$CO_{2}\ (4.4\ gm)$$

Hence, it proves the law of conservation of mass.
Question 2
1 / -0

A solution is prepared by dissolving 5.64 g of glucose in 60 g of water. Calculate the mass percent of glucose.

A
8.59%

B
6.85%

C
9.34%

D
3.59%

Solution

Mass percent of glucose $$=\displaystyle \frac{mass\ of\ glucose}{mass\ of\ solution} \times 100$$
$$=\displaystyle \frac{5.64}{(5.64 + 60)} \times 100 = $$ 8.59%
Question 3
1 / -0

18 carat gold contains ?

A
18 % gold

B
24 % gold

C
75 % gold

D
60 % gold

Solution

Caratage is the measurement of purity of gold alloyed with other metals. 24 carat is pure gold with no other metals. Low caratages contain less gold. 18 carat gold contain $$75\%$$ of gold & $$25\%$$ of other metals often, Copper/Silk.
Question 4
1 / -0

Active mass of 6% solution of compound X is 2. Molecular weight of X would be :

A
$$6$$

B
$$30$$

C
$$60$$

D
$$90$$

Solution

Active mass of 2 means 2 moles of X in one litre solution.
$$\because 100\quad ml\quad solution\quad contains\quad 6\quad gm\quad of\quad X\\ \therefore 1000\quad ml\quad solution\quad contains\quad \frac { 6\times\ 1000 }{ 100 } =\quad 60\quad gm\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2\quad moles\quad =\quad 60\quad gm\quad X\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 1\quad mole\quad =\quad 30\quad gm\quad X\\ Hence\quad molecular\quad weight\quad of\quad X\quad is\quad 30.$$
Question 5
1 / -0

The main constituent of LPG is butane. Then
(A) Butane can be liquefied easily under high pressure
(B) Butane is liquefied by chemically reacting with ethane and propane.

A
Only A is true

B
Only B is true

C
Both A and B are true

D
Both A and B are false

Solution

$$\text{Butane can be liquefied easily under high pressure. Ex: In LPG cylinder.}$$
$$\text{Burane cannot be liquefied by any chemical reaction.}$$
Question 6
1 / -0

Two gases A and B are taken in same volume containers under similar conditions of temperature and pressure. In container A, there are '2N' molecules of gas A. The number molecules does container B have :

A
N

B
0.5N

C
2N

D
4N

Solution

1. According to Avogadro's law, the equal volume of gases under a similar condition of pressure and temperature contain equal no. of molecules.
Therefore the number of molecules in container B will also be 2N.

2. As we know, one mole of each gas occupies 22.4-litre volume and one mole of gas contains N$$_A$$ molecules at STP.

So at the same temperature and pressure, the equal volume of gas contains equal no of molecules.

Therefore the number of molecules in container B will also be 2N.

Therefore, the option is C.
Question 7
1 / -0

Gram molar volume for a gas is always considered at ___ conditions.

A
NTP

B
STP

C
variable temperature

D
none of these

Solution

Since volume is a dependent variable and it depends on pressure and temperature always so temperature and pressure should be specified when volume of gas is considered. For this certain standard conditions are taken as 273 K and 1 atm pressure.
And at STP, molar volume is 22.4 litre.
Question 8
1 / -0

Assuming that the all volume is measured at the same temperature and pressure, state the volume ratios of the reactants and products for the following gaseous reactions. Nitrogen reacting with oxygen to form nitrogen (III) oxide.

A
1 : 3 : 2

B
2 : 4 : 2

C
2 : 2: 1

D
2 : 3 : 2

Solution

The equation for the reaction between nitrogen and oxygen to form Nitrogen (lll) oxide is written as:
$$2N_{2} + 3O_{2} \rightarrow 2N_{2}O_{3}$$
Therefore the volume ratio of reactants and products $$= 2 : 3 : 2$$
Question 9
1 / -0

Water-gas is an equimolar mixture of hydrogen and carbon monoxide. If 250 litres of water gas is burnt in air. Assuming both reactants and products are at the same temperature and pressure, the volume of $$CO_2$$ produced is:

$$2CO + O_2 \rightarrow 2CO_2;$$

$$2H_2 + O_2 \rightarrow 2H_2O$$

A
125 litres

B
62.5 litres

C
250 litres

D
500 litres

Solution

$$2CO + O_2 = 2CO_2;$$

$$2H_2 + O_2 = 2H_2O$$

Given, $$CO + H_2 = 250 L$$

Since $$CO$$ and $$H_2$$ are equimolar, so each has vol 125 L (Given total volume is 250 L.)

The equation shows that :

2 L of $$CO$$ gives 2 L of $$CO_2$$

So, 125 L of $$CO$$ gives 125 L of $$CO_2$$

Hence, option (a) is correct.
Question 10
1 / -0

Water gas is an equi-molar mixture of hydrogen and carbon monoxide. If 250 litres of water gas is burnt in air. Assuming both reactants and products are at same temperature and pressure, the volume of $$O_2$$ required is _________.
$$2CO + O_2 \rightarrow 2CO_2; 2H_2 + O_2 \rightarrow 2H_2O$$

A
250 litres

B
125 litres

C
75 litres

D
500 litres

Solution

Equation for combustion of water gas in air is:

$$2CO + O_{2} \rightarrow 2CO_{2}$$ & $$2H_{2} + O_{2} \rightarrow 2H_{2}O$$

For combustion of $$1$$ mole of water gas ($$CO + H_{2}$$) , $$1/2$$ mole of $$O_{2}$$ is required.

Therefore, for combustion of $$250L$$ of water gas, amount of oxygen required $$= \dfrac{250}{2}= 125L$$ of $$O_2$$ is required.

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Some Basic Concepts of Chemistry Test - 21

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Some Basic Concepts of Chemistry Test - 21

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Question 1

law of constant composition

law of multiple proportion

law of reciprocal proportion

law of conservation of mass

Solution

Question 2

8.59%

6.85%

9.34%

3.59%

Solution

Question 3

18 % gold

24 % gold

75 % gold

60 % gold

Solution

Question 4

$$6$$

$$30$$

$$60$$

$$90$$

Solution

Question 5

Only A is true

Only B is true

Both A and B are true

Both A and B are false

Solution

Question 6

N

0.5N

2N

4N

Solution

Question 7

NTP

STP

variable temperature

none of these

Solution

Question 8

1 : 3 : 2

2 : 4 : 2

2 : 2: 1

2 : 3 : 2

Solution

Question 9

125 litres

62.5 litres

250 litres

500 litres

Solution

Question 10

250 litres

125 litres

75 litres

500 litres

Solution

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