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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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