NEET CHEMISTRY

DAILY CHEMISTRY TEST QUESTIONS

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1 Surface Chemistry

526.3 mL of 0.5 m HCl is shaken with 0.5 g of activated charcoal and filtered. The concentration of the filtrate is reduced to 0.4 m. The amount of adsorption of HCl per unit mass of adsorbent is

1) 3
2) 6
3) 8
4) 4

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Since the solution is dilute, molality can be approximated to molarity.
Due to adsorption, the molarity changes from 0.5 M to 0.4 M.

Hence, the number of moles of HCl adsorbed =(0.5−0.4)× (526.3 /1000) =0.05263 moles.

The mass of HCl adsorbed x =0.05263×36.5=1.9209 g
The mass of charcoal (m) is 0.5 g.

The amount of adsorption (x? /m) = 1.9209 /0.5 ≈4

2 Stoichiometry

600 m of a mixture of O₃and O₂ weight 1 g at NTP. Calculate the volume of ozone in the mixture.

1) 400 ml
2) 600 ml
3) 200 ml
4) 100 ml

Option 1 Solution:

Solution

The molecular mass of oxygen= 32

The molecular mass of ozone=48

Therefore,

Let volume of $${ O }_{ 2 }=x\quad ml$$

volume of $${ O }_{ 3 }=600- x\ ml$$

At STP, wt of x ml of $${ O }_{ 2 }=\dfrac { 32 \times x }{ 22400 } gm$$

wt of $$600-x\ ml\ { O }_{ 3 }=\dfrac { 48\times (600-x) }{ 22400 } gm$$

Hence,$$\dfrac { 32\ \times x }{ 22400 } + \dfrac { 48\times (600-x) }{ 22400 } =1\\$$

Therefore, volume of ozone in the mixture is $$400\ ml$$

Option 2

Option 3

Option 4

3 Solid State

If the lattice parameter of Si is 5.43 Å and the mass of Si atom is 28.08×1.66×10⁻²⁷ kg, the density of silicon in kg m⁻³ is :

(Given: Silicon has diamond cubic structure.)

1) 2330
2) 1115
3) 3445
4) 1673

Option 1 Solution:

Solution

SInce Si has diamond cubic structure, its unit cell contains $$8$$ atoms.
Density $$\displaystyle = \dfrac {\text{Mass}}{\text{Volume}} $$
Mass $$\displaystyle = 8 \times 28.08 \times 1.66 \times 10^{-27} = 3.73 \times 10^{-25} $$ kg
Volume $$\displaystyle = (5.43 \times 10^{-10})^3 = 1.6 \times 10^{-28} \: $$ m$$^3$$
Density $$\displaystyle = \frac {3.73 \times 10^{-25} }{ 1.6 \times 10^{-28}} = 2330$$ kg m$$^{-3}$$

Option 2

Option 3

Option 4

4 Alcohols Phenols and Ethers

The final product of oxidation of $$2-propanol$$ with hot con. $$HNO_{3}$$ is

1) Propanal
2) Ethanoic acid
3) Propanone
4) Ethanol

Option 1

Option 2 Solution:

Solution

$$\underset {propan-2-ol}{CH_{3}CHOHCH_{3}} \xrightarrow {[O]} \underset {acetone}{CH_{3}COCH_{3}}\xrightarrow {[O]} \underset {\text {acetic acid or ethanoic acid}}{CH_{3}COOH} + CO_{2} + H_{2}O$$.

Option 3

Option 4

5 Hydrogen

One mole of N_2?H_4?loses 10 moles of electrons to form a new compound Y. Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in compound Y?

[There is no change in the oxidation state of hydrogen.]

1) −1
2) −3?
3) +3
4) +5

Option 1

Option 2

Option 3 Solution:

Solution

One Mole of N₂?H₄? loses 10 moles of e− ? Y

The oxidation state of H remains unchanged and all the Nitrogen" appears in Y.

∴ e− must be lost from nitrogen,

1 mole of N₂?H₄? contains 2 moles of nitrogen.

?2 moles of N loses 10 moles of e−.

∴1 mole loses 5 moles of e−

Initial oxidation state of N in N2?H4?=−2

when it loses 5 moles of e−

Final oxidation state of each N=−2+5=+3.

(since it loses e−, therefore it must acquire positive charge).

Option 4

6 Aldehydes Ketones and Carboxylic Acids

Which of the following will form when calcium acetate is distilled?

1) Formaldehyde

2) Ethanal

3) Propanal

4) Acetone

Option 1

Option 2

Option 3 Solution:

Option 4

7 S � Block Elements

The metal oxides of alkaline earth metal can be formed by heating corresponding carbonates except:

1) BeO
2) MgO
3) CaO
4) none

Option 1 Solution:

Solution

The metal oxides of alkaline earth metal can be formed by heating corresponding carbonates except BeOBeO due to small size of beryllium ion.BeO is covalent and has polymeric nature.

Option 2

Option 3

Option 4

8 Solid State

In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids are occupied by divalent (X⁻) ions, while one-half of the octahedral voids are occupied by trivalent ions (Y⁺³), then the formula of the oxide is:

1) XY₂O₄
2) X₂YO_4???????
3) X₄Y₅O₁₀
4) X₅Y₄O₁₀

Option 1

Option 2

Option 3 Solution:

Solution

In a cubic packed structure,

$$O^{2-}$$ present in ccp=4 $$O^{2-}$$ ions

Tetrahedral voids=8

Octahedral voids=4

Thus, $$X^{-}$$ ions=$$\cfrac{8}{5}$$

$$Y^{+3}$$ ions=$$\cfrac{4}{2}=2$$

So, the molecular formula is $$X_{\cfrac{8}{5}}Y_{2}O_{4}$$.

In simplest form, $$X_{4}Y_{5}O_{10}$$

Option 4

9 Chemistry in Everyday Life

When sodium stearate CH3(CH2)16COONa, is added to hard water the formula of one possible precipitate is:

1) (CH₃(CH₂)₁₆COO)₂Ca

2) (CH₃(CH₂)₁₆COO)₂Ba

3) (CH₃(CH₂)₁₆COO)₂Cd

4) CH₃(CHv)₁₆COOK

Option 1 Solution:

Solution

When sodium stearate CH₃(CH₂)₁₆COONa, is added to hard water, calcium of hard water reacts with the soap and replaces its sodium and forms calcium precipitate, (CH₃(CH₂)₁₆COO)₂Ca

Option 2

Option 3

Option 4

10 Organic Compounds Containing Nitrogen

Hydrolysis of benzonitrile by dilute HCl yields

1. aniline
2. benzoldelyde
3. benzoic acid
4. benzaunide

Option 1

Option 2

Option 3 Solution:

Option 4

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