Some Basic Concepts of Chemistry Test

Result

Some Basic Concepts of Chemistry Test - 58

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The percentage loss in weight after heating a pure sample of potassium chlorate (mol. $$wt.=122.5$$) will be:

A
12.25

B
24.50

C
39.18

D
19.6

Solution

$$2KClO_3\longrightarrow 2KCl+3O_2$$
$$2\times 122.5$$ $$g$$ $$3\times 32$$ $$g$$ $$O_2$$
$$\implies 2\times 122.5$$ $$g$$ $$KClO_3$$ show loss of $$3\times 32$$ $$g$$ of $$O_2$$
$$100$$ $$g$$ $$KClO_3$$ shows weight loss of $$\cfrac {3\times 32\times 100}{2\times 122.5}=39.17$$
$$\implies \%$$ Loss of weight $$=39.17$$
Question 2
1 / -0

The carbonate of a metal is isomorphous with $$MgCO_{3}$$ and contains $$6.091\%$$ of carbon. Atomic weight of the metal is nearly:

A
$$48$$

B
$$68.5$$

C
$$137$$

D
$$120$$

Solution

The molecular formula of metal carbonate is $${MCO}_{3}$$ as it is isomorphous with $${MgCO}_{3}$$.
Let the mass of metal be $$M$$. Molecular weight of this carbonate $$= (M+12+48)$$. Mass percentage of carbon in this compound $$= 6.091%$$
$$\dfrac{6.091}{100}\times(M+12+48) = 12$$
$$M = \dfrac{12\times100}{6.091} - 60$$
$$M = 137$$

Atomic wt. of the element is 137.01 amu which is Barium and the metal carbonate is BaCO$$_3$$

Hence, the correct option is $$C$$
Question 3
1 / -0

$$10\ { dm }^{ 3 }$$ of $${ N }_{ 2 }$$ gas and $$10\ { dm }^{ 3 }$$ of gas X at the same temperature contain the same number of molecules. The gas X is

A
$$CO$$

B
$${ CO }_{ 2 }$$

C
$${ H }_{ 2 }$$

D
$$NO$$

Solution

$$CO$$

$$C-12,O-16$$

$$12+16=28$$

molar mass of $$N_2$$

$$N_2=14+4=28$$

Hence $$CO$$ is the answer
Question 4
1 / -0

If $$N _ { A }$$ is Avogadro's number, then calculate the number of electrons in $$4.2\ g$$ of $$N ^ { 3- }$$.

A
$$2.1 N _ { A }$$

B
$$3 N _ { A }$$

C
$$6 N _ { A }$$

D
$$ N _ { A }$$

Solution

$$N^{3-}$$ contains 10 electrons.

Number of moles $$= \dfrac {4.2}{14} $$

$$=0.3\ moles$$

No of electrons in $$0.3\ moles = 0.3\times N_A\times 10$$

$$=3\ N_A$$

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

A solution containing $$500 g$$ of a protein per litre is isotonic with a solution containing $$3.42 g$$ of sucrose per litre. The molecular mass of protein is:

A
$$5$$

B
$$146$$

C
$$34200$$

D
$$50000$$

Solution

Two solutions are said to isotonic when they are at the same concentrations.

The molar mass of sucrose $$=342$$

$$C_1 =$$ Concentration of protein
$$C_2 =$$ Concentration of sucrose

Since, they are isotonic, $$C_1 = C_2$$

$$\cfrac {500}{\text{Molar mass of protein}} = \cfrac {3.42}{342}$$

$$\implies $$ Molar mass of protein $$=500\times 100=50000$$ $$g$$
Question 6
1 / -0

The Statue of Liberty is made of $$2.0\times 10^5$$ lbs of copper sheets bolted to a framework $$(1 lb =454 g)$$. How many atoms of copper are there on the statue?

(Atomic weight: $$Cu=63.5$$).

A
$$2.1\times 10^{27}$$

B
$$8.6\times 10^{29}$$

C
$$4.3\times 10^{26}$$

D
$$8.6\times 10^{26}$$

Solution

Given,
mass of copper $$= 2 \times 10^5 lbs$$

$$= 2 \times 10^5 \times 454 g$$

Molecular wt. of copper $$= 63.5 g/moles$$

$$\therefore$$ No. of moles of copper $$= \dfrac{2 \times 10^5 \times 454}{63.5}$$

Thus, no. of atoms of copper on the statue

$$= \dfrac{2 \times 10^5 \times 454}{63.5} \times 6.023 \times 10^{23}$$

$$= 8.6 \times 10^{29}$$ atoms

Hence, the correct option is (B)
Question 7
1 / -0

The weight of $$H_{2}C_{2}O_{4},2H_{2}O$$ required to prepare 500 ml of 0.2 N solution is:

A
1.26 g

B
6.3 g

C
1.575 g

D
3.15 g

Solution
Question 8
1 / -0

Number of $$HCl$$ molecules present in $$10$$ mL of $$0.1 N$$ $$HCl$$ solution is

A
$$6.022 \times 10^{23}$$

B
$$6.022 \times 10^{22}$$

C
$$6.022 \times 10^{21}$$

D
$$6.022 \times 10^{20}$$
Question 9
1 / -0

One gram molecule of any gas at $$NTP$$ occupies $$22.4\ L$$. This fact was derived from:

A
Dalton's theory

B
Avogardro's hypothesis

C
Berzelius hypothesis

D
Law of gaseous volume

Solution

According to Avogadro's law, 1 mole of every gas occupies 22.4L at $$NTP$$.
From ideal gas equation,
$$PV=nRT$$. . . . . . .(1)
At $$NTP$$,
$$T=273K$$
$$P=1atm$$
$$n=1$$
$$R=0.0821atm L/K mol$$
from equation (1),
$$V=\dfrac{nRT}{P}$$
$$V=\dfrac{1\times 0.0821\times 273}{1}=22.4L$$
The correct option is B.
Question 10
1 / -0

A compound contains $$28\%$$ Nitrogen, by mass. The possible molecular mass of the compounds is/are?

A
$$100$$

B
$$125$$

C
$$150$$

D
$$140$$

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 58

Result

Some Basic Concepts of Chemistry Test - 58

Score

-

Rank

-

SHARING IS CARING

Question 1

12.25

24.50

39.18

19.6

Solution

Question 2

$$48$$

$$68.5$$

$$137$$

$$120$$

Solution

Question 3

$$CO$$

$${ CO }_{ 2 }$$

$${ H }_{ 2 }$$

$$NO$$

Solution

Question 4

$$2.1 N _ { A }$$

$$3 N _ { A }$$

$$6 N _ { A }$$

$$ N _ { A }$$

Solution

Question 5

$$5$$

$$146$$

$$34200$$

$$50000$$

Solution

Question 6

$$2.1\times 10^{27}$$

$$8.6\times 10^{29}$$

$$4.3\times 10^{26}$$

$$8.6\times 10^{26}$$

Solution

Question 7

1.26 g

6.3 g

1.575 g

3.15 g

Solution

Question 8

$$6.022 \times 10^{23}$$

$$6.022 \times 10^{22}$$

$$6.022 \times 10^{21}$$

$$6.022 \times 10^{20}$$

Question 9

Dalton's theory

Avogardro's hypothesis

Berzelius hypothesis

Law of gaseous volume

Solution

Question 10

$$100$$

$$125$$

$$150$$

$$140$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.