Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 58

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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

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 58

Result

Some Basic Concepts of Chemistry Test - 58

Score

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Rank

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SHARING IS CARING

Question 1

12.25

24.50

39.18

19.6

Solution

Question 2

484848

68.568.568.5

137137137

120120120

Solution

Question 3

COCOCO

CO2{ CO }_{ 2 }CO2​

H2{ H }_{ 2 }H2​

NONONO

Solution

Question 4

2.1NA2.1 N _ { A }2.1NA​

3NA3 N _ { A }3NA​

6NA6 N _ { A }6NA​

NA N _ { A }NA​

Solution

Question 5

555

146146146

342003420034200

500005000050000

Solution

Question 6

2.1×10272.1\times 10^{27}2.1×1027

8.6×10298.6\times 10^{29}8.6×1029

4.3×10264.3\times 10^{26}4.3×1026

8.6×10268.6\times 10^{26}8.6×1026

Solution

Question 7

1.26 g

6.3 g

1.575 g

3.15 g

Solution

Question 8

6.022×10236.022 \times 10^{23}6.022×1023

6.022×10226.022 \times 10^{22}6.022×1022

6.022×10216.022 \times 10^{21}6.022×1021

6.022×10206.022 \times 10^{20}6.022×1020

Question 9

Dalton's theory

Avogardro's hypothesis

Berzelius hypothesis

Law of gaseous volume

Solution

Question 10

100100100

125125125

150150150

140140140

Solution

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$48$

$68.5$

$137$

$120$

$CO$

${ CO }_{ 2 }$

${ H }_{ 2 }$

$NO$

$2.1 N _ { A }$

$3 N _ { A }$

$6 N _ { A }$

$N _ { A }$

$5$

$146$

$34200$

$50000$

$2.1\times 10^{27}$

$8.6\times 10^{29}$

$4.3\times 10^{26}$

$8.6\times 10^{26}$

$6.022 \times 10^{23}$

$6.022 \times 10^{22}$

$6.022 \times 10^{21}$

$6.022 \times 10^{20}$

$100$

$125$

$150$

$140$