Download CBSE Sample Paper 2025-26 for class 12th to 8th

Some Basic Concepts of Chemistry Test - 75

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Some Basic Concepts of Chemistry Test - 75

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Weekly Quiz Competition

Question 1
1 / -0

$3 g$ of activated charcoal was added to $50 mL$ of acetic acid solution ( $0.06 N$ ) in a flask. After an hour it was filtered and the strength of the filtrate was found to be $0.042 N$ . The amount of acetic acid absorbed (per gram of charcoal) is ?

A
$18 mg$

B
$36 mg$

C
$42 mg$

D
$54 mg$

Solution

Given:

The initial mass of charcoal $= 3g$

Vol. of acetic acid = $50mL$ Normality $= 0.06N$

Strength of filtrate after reaction $= 0.042N$

We know,

Molarity of acetic acid solⁿ $= \cfrac{{Wt/M}}{{Vol\ of\ so{l^n}}}$

$\Rightarrow Wt = 0.06 \times 60 \times 0.05 = 0.180g$

Also, normality after reaction, $0.042 = \cfrac{W}{{60 \times 0.05}}$

$W = 0.126g$

Change in mass $= 0.180 - 0.126 = 0.054g = 54mg$

Amount of charcoal available $= 3g$

Amount of acetic acid absorbed $= \cfrac{{54mg}}{{3.0g}} \times 1.0g = 18g$
Correct option is A.
Question 2
1 / -0

A substance contains 3.4% $S$ . if it contain two atom of sulphur per molecule then minimum molecular weight of substance will be:

A
188.2

B
94.1

C
1882

D
282.3

Solution

According to question-
$3.4g S=100g$ insulin

$\therefore 32g\ S= \dfrac{100\times 32}{3.4}$

$\Rightarrow 941.176$
For two atoms of $S$
$\Rightarrow 941.176\times 2 = 1882.4$

Here correct unit is $\dfrac{18882.4}{10}=188.2$

Insulin must contain at least Two atom of S in its one molecule.

Hence $(A)$ is the correct answer.
Question 3
1 / -0

The number of $H_{3}O^{+}$ ions present in 10ml of water at $25^{\circ}C$ is:

A
$6.023 \times 10^{-14}$

B
$6.023 \times 10^{14}$

C
$6.023 \times 10^{-19}$

D
$6.023 \times 10^{19}$

Solution

No. of hydronium ions present in $10$ ml water at ${ 25 }^{ 0 }C$ .

$pH=7$ at ${ 25 }^{ 0 }C$

$\left[ { H }^{ + } \right] ={ 10 }^{ -7 }M$

No. of moles $=$ Concentration $\times$ Volume

$=\left( { 10 }^{ -7 } \right) \left( 10 \right) \times { 10 }^{ -3 }$

$={ 10 }^{ -9 }$ moles

1 mole has $6.023\times { 10 }^{ 23 }$ ions of hydronium.

$\Rightarrow \quad 6.023\times { 10 }^{ 23 }\times { 10 }^{ -9 }$ ions in $10$ ml water

$\Rightarrow 6.023\times { 10 }^{ 14 }$ ions are there in $10$ ml water at ${ 25 }^{ 0 }C$ .

Hence, option B is correct.
Question 4
1 / -0

The volume of $0.1\ M\ H_2SO_4$ required to neutralize 30 mL of $2.0\ M\ NaOH$ is:

A
100 mL

B
300 mL

C
400 mL

D
200 mL

Solution

Solution:- (B) $300 \; mL$
Let $V$ be the volume of ${H}_{2}S{O}_{4}$ required.
Given:-
${M}_{{H}_{2}S{O}_{4}} = 0.1 M$
${M}_{NaOH} = 2 M$
${V}_{NaOH} = 30 \; mL$
Now, as we know that
${N}_{1} {V}_{1} = {N}_{2} {V}_{2}$
$\because N = n \times M$
Whereas $n$ is the valence factor
Therefore,
${n}_{1} {M}_{1} {V}_{1} = {n}_{2} {M}_{2} {V}_{2}$
For ${H}_{2}S{O}_{4}; \; n = 2$
For $NaOH; \; n = 1$
Therefore,
$2 \times 0.1 \times V = 1 \times 2 \times 30$
$\Rightarrow V = \cfrac{60}{0.2}$
$\Rightarrow V = 300 \; mL$
Hence $300 \; mL$ of ${H}_{2}S{O}_{4}$ is required to neutralize $30 \; mL$ of $2 M \; NaOH$ .
Question 5
1 / -0

Two oxide of metal contain 27.6% and 30% oxygen respectively. If the formula of first oxide is $M_3O_4$ then formula of second oxide is -

A
$MO$

B
$M_2O$

C
$M_2O_3$

D
$MO_2$
Question 6
1 / -0

$81gm$ mixture of $MgC{O_3}\left( s \right)$ $N{H_2}COON{H_4}\left( s \right)$ in equimolar ratio is heated in open to constant mass. If vapour density of gaseous mixture evolved was found to be $\cfrac{61}{4}$ , then find the mole % of $MgCO_3$ in original sample.

Given that: $MgC{O_3}\left( s \right)\xrightarrow{\Delta }MgO\left( s \right) + C{O_2}\left( g \right)$

${H_2}COON{H_4}\left( s \right)\xrightarrow{\Delta }2N{H_3}\left( g \right) + C{O_2}\left( g \right)N$

A
$50\%$

B
$51.9\%$

C
$70.9\%$

D
$59.1\%$
Question 7
1 / -0

Substance
Boiling point ${ ( }^{ 0 }C)$
Melting point ${ ( }^{ 0 }C)$
A -183 -219
B 445 119
C 78 -15

Point out the physical states of A, B, and C at room temperature $({ 30 }^{ 0 }C)$ .

A
Gas, solid, liquid

B
Gas, liquid, solid

C
Liquid, solid, gas

D
Solid, liquid, gas

Solution

$(i)$ If Boiling point $>$ Room Temperature $>$ Melting Point
Then the physical state is liquid.

$(ii)$ If Room temperature $>$ Boiling point
Then the physical state is gas.

$(iii)$ If Room temperature $<$ Melting point
Then the physical state is solid.

Therefore,
for $A\Rightarrow$ Physical state is gas.
For $B\Rightarrow$ Physical state is solid.
For $C\Rightarrow$ Physical state is liquid.
Question 8
1 / -0

A mixture of ${ C }_{ 2 }{ H }_{ 2 }$ AND ${ C }_{ 3 }{ H }_{ 8 }$ occupied a certain volume at 80 mm Hg . The mixture wsa completely burnt to ${ CO }_{ 2 }$ and ${ H }_{ 2 }O$ (l). When the pressure of ${ CO }_{ 2 }$ was found to be 230 mm Hg at the same temperature and volume, the mole fraction of ${ C }_{ 3 }{ H }_{ 8 }$ in the mixture is

A
0.125

B
0.875

C
0.6

D
0.8

Solution
Question 9
1 / -0

Vapour density of the equilibrium mixture of $NO_2$ and $N_2O_4$ is found to be $40$ for the equilibrium. Calculate percentage of $NO_2$ in the mixture: $N_2O_4 \leftrightharpoons 2 NO_2$

A
$10$ %

B
$5$ %

C
$26.08$ %

D
None of these

Solution
Question 10
1 / -0

0.1 g of a solution containing ${ Na }_{ 2 }$ ${ Co }_{ 3 }$ and ${ NaHCO }_{ 3 }$ requires 10 mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator.Wt.% of ${ Na }_{ 2 }$ ${ Co }_{ 3 }$ in the mixture is

A
25

B
32

C
50

D
10.6

Solution