Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 70

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Question 1
1 / -0

Maximum percentage of chlorine is in :

A
Chloral

B
Pyrene

C
Ethylidene chloride

D
PVC

Solution

$\text{Percentage of chlorine} = \dfrac{\text{Mass of chlorine in the compound}}{\text{Mass of compound}}\times 100$

Chloral is $Cl_3CCHO$ .
Percentage $=\dfrac{106.5}{147.5}\times100=72.2$

Pyrene is $CCl_4$ .
Percentage $=\dfrac{142}{154}\times 100=92.20$

Ethylidene chloride is $C_2H_2Cl_2$ .
Percentage $=\dfrac{71}{97}\times100=73.19$

PVC is $[CH_2CHCl]_n$ .
Percentage $=\dfrac{35.5}{62.5}\times100=56.8$

Therefore, the maximum percentage of chlorine is present in pyrene.
Question 2
1 / -0

Avogadro's law shows the relationship between which two variables?

A
Volume and number of moles

B
Pressure and number of moles

C
Volume and pressure

D
Temperature and pressure

E
Temperature and number of moles

Solution

(A) : Volume and number of moles, : at constant pressure.
Avogadro Law: At the same condition of temperature and pressure the volume of the gas is directly proportional to the number of the moles.
Question 3
1 / -0

Statement I : At STP, 22.4 liters of He will have the same volume as one mole of $\displaystyle { H }_{ 2 }$ (assume ideal gases).
Statement II : One mole or 22.4 liters of any gas at STP will have the same mass.

A
true, false

B
false, true

C
true, true, correct explanation

D
true, true, not correct explanation

Solution

Avogadro’s law states that under the same conditions of temperature and pressure, equal volumes of different gases contain an equal number of molecules. This empirical relation can be derived from the kinetic theory of gases under the assumption of a perfect (ideal) gas. The law is approximately valid for real gases at sufficiently low pressures and high temperatures.
At STP, all gas have same volume for 1 mol of gas and that volume is always equal to the 22.4 L.
Question 4
1 / -0

Natural abundances of $^{12}C$ and $^{13}C$ isotopes of carbon are $99\%$ and $1\%$ , respectively. Assuming they only contribute to the mol. wt. of $C_2F_4$ , the percentage of $C_2F_4$ having a molecular mass of $101$ is:

A
$1.00$

B
$98$

C
$0.198$

D
$99$

Solution

The molecular mass of $\displaystyle C_2F_4$ containing C-12 is $\displaystyle 2(12)+4(19)=100$ g/mol
The molecular mass of $\displaystyle C_2F_4$ containing one C-12 and one C-13 is $\displaystyle 12+13+4(19)=101$ g/mol
Natural abundances of C-12 and C-13 isotopes of carbon are 99% and 1%, respectively.
100 $\displaystyle C_2F_4$ molecules will have 200 C atoms out of whic 1% or 2 C atoms will be C-13. Assuming that these two C-13 atoms will be in different $\displaystyle C_2F_4$ molecules, the percentage of $\displaystyle C_2F_4$ having a molecular mass of 101 is $\displaystyle \dfrac{2}{200} \times 100 = 1$ %
Question 5
1 / -0

1 g of silver gets distributed between $10\ cm^3$ of molten zinc and $100\ cm^3$ of molten lead at $800^o C$ . The percentage of silver in the zinc layer in approximately:

A
89%

B
91%

C
97%

D
94%

Solution

Concentration of $Ag$ is zinc $=\dfrac{x}{10}$

$\therefore$ concentration of $Ag$ in $Pb$ $=\dfrac{1-x}{100}$
$\dfrac{\dfrac{x}{10}}{\dfrac{1-x}{100}}=300$
On solving, we get $x=97\%$
Question 6
1 / -0

What is the percentage of sulphur in sulphuric acid $(H_2SO_4)$ ?

A
32%

B
32.45%

C
32.65%

D
32.55%

Solution

Molecular weight of ${ H }_{ 2 }{ SO }_{ 4 }=2+32+64=98g$
$98g$ contains $32gm$ of $S$
$\therefore$ % of $S$ in ${ H }_{ 2 }{ SO }_{ 4 }$ $=\dfrac { 32 }{ 98 } \times 100=32.65$ %
Question 7
1 / -0

Two vessels of volumes $16.4\ L$ and $5\ L$ contains two ideal gases of molecular existence at the respective temperature of $27^{\circ}C$ and $227^{\circ}C$ and exert $1.5$ and $4.1$ atm, respectively. The ratio of the number of molecules of the former to that of the later is:

A
$2$

B
$1$

C
$\dfrac {1}{2}$

D
$\dfrac {1}{3}$

E
$3$

Solution

From the ideal gas equation we have,

$\dfrac {p_{1}V_{1}}{n_{1}T_{1}} = \dfrac {p_{2}V_{2}}{n_{2}T_{2}}=R$

$\dfrac {n_{1}}{n_{2}} = \dfrac {p_{1}V_{1}T_{2}}{p_{2}V_{2}T_1} = \dfrac {1.5\times 16.4\times 500}{4.1\times 5\times 300} = 2$

$\therefore n_{1} : n_{2} = 2 : 1$ .
Question 8
1 / -0

The total number of electrons present in $18 \ mL$ water (density $1g/mL$ ) is:

A
$6.023\times { 10 }^{ 23 }$

B
$6.023\times { 10 }^{ 24 }\quad$

C
$6.023\times { 10 }^{ 25 }$

D
$6.023\times { 10 }^{ 21 }\quad$

Solution

Mass $=18g$ ( $\because m=v\times d$ )

Number of molecules of ${H}_{2}O$ in $18g$ mass

$\quad =6.023\times { 10 }^{ 23 }$

Number of electrons in $18g$ water

$=6.023\times { 10 }^{ 23 }\times 10$

$=6.023\times { 10 }^{ 24 }\quad$

Hence, the correct option is $\text{B}$
Question 9
1 / -0

The density of a liquid is 1.2 g/mL. That are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70):

A
$\frac{1.2}{35} N_A$

B
$\left (\frac{1}{35} N_A \right )^2$

C
$\frac{1.2}{(35)^2} N_A$

D
$1.2 N_A$

Solution

Volume of one drop $=\frac{2}{35}\, mL$

Moles in one drop $=\frac{2 \times 1.2}{35 \times 70}=\frac{1.2}{(35)^2}\ mol$

Number of molecules in one drop $=\frac{1.2}{(35)^2}\times N_A$
Question 10
1 / -0

20 ml of $H_2O_2$ after acidification with dil $H_2SO_4$ required 30 ml of N/2 $KMnO_4$ for complete oxidation .Calculate the $\%$ of $H_2O_2$ in g/lit.

A
10.75 g/lit

B
11.75 g/lit

C
12.75 g/lit

D
13.75 g/lit

Solution

Given,
Volume of $H_2O_2=20ml$
Volume of $KMnO_4=30ml$
Normality of $KMnO_4=\cfrac {1}{2}N$
Now,
We know at equivalence
Gram equivalent Weight of $H_2O_2$ = Gram equivalent weight of $KMnO_4$
i.e. $\cfrac {Wt_{H_2O_2}}{Eq.Wt._{H_2O_2}}=N_{KMnO_4}\times V_{KMnO_4}$
$=\cfrac {1}{2}N\times 0.03L$
$=0.015$
$\Rightarrow$ Weight of $H_2O_2=0.015\times Eq.Wt of H_2O_2$
$=0.015\times \cfrac {34}{2}$
$=0.255$
Now, Weight/Volume % of $H_2O_2=\cfrac {Weight\quad of\quad H_2O_2}{Volume\quad of\quad H_2O_2}\times 100$
$=\cfrac {0.255}{20\times 10^{-3}}\times 100$
$=12.75$ %

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 70

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Some Basic Concepts of Chemistry Test - 70

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SHARING IS CARING

Question 1

Chloral

Pyrene

Ethylidene chloride

PVC

Solution

Question 2

Volume and number of moles

Pressure and number of moles

Volume and pressure

Temperature and pressure

Temperature and number of moles

Solution

Question 3

true, false

false, true

true, true, correct explanation

true, true, not correct explanation

Solution

Question 4

1.001.001.00

989898

0.1980.1980.198

999999

Solution

Question 5

89%

91%

97%

94%

Solution

Question 6

32%

32.45%

32.65%

32.55%

Solution

Question 7

222

111

12\dfrac {1}{2}21​

13\dfrac {1}{3}31​

333

Solution

Question 8

6.023×10236.023\times { 10 }^{ 23 }6.023×1023

6.023×10246.023\times { 10 }^{ 24 }\quad 6.023×1024

6.023×10256.023\times { 10 }^{ 25 }6.023×1025

6.023×10216.023\times { 10 }^{ 21 }\quad 6.023×1021

Solution

Question 9

1.235NA\frac{1.2}{35} N_A351.2​NA​

(135NA)2\left (\frac{1}{35} N_A \right )^2(351​NA​)2

1.2(35)2NA\frac{1.2}{(35)^2} N_A(35)21.2​NA​

1.2NA1.2 N_A1.2NA​

Solution

Question 10

10.75 g/lit

11.75 g/lit

12.75 g/lit

13.75 g/lit

Solution

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$1.00$

$98$

$0.198$

$99$

$2$

$1$

$\dfrac {1}{2}$

$\dfrac {1}{3}$

$3$

$6.023\times { 10 }^{ 23 }$

$6.023\times { 10 }^{ 24 }\quad$

$6.023\times { 10 }^{ 25 }$

$6.023\times { 10 }^{ 21 }\quad$

$\frac{1.2}{35} N_A$

$\left (\frac{1}{35} N_A \right )^2$

$\frac{1.2}{(35)^2} N_A$

$1.2 N_A$