NEET CHEMISTRY question bank, MCQ CBSE Notes, Lectures

1 Biomolecules

Which one of the following chars when warmed with concentrated sulphuric acid

1. hydrocarbon
2. carbohydrate
3. protein
4. fat

Option 1

Option 2 Solution:

Option 3

Option 4

2 Ionic Equilibrium

Phosphoric acid is commonly supplied as an 85% solution. What mass of this solution would be required to prepare 600.0 mL of a 2.80 M phosphoric acid solution?

1) 104 g
2) 280 g
3) 194 g
4) none

Option 1

Option 2

Option 3 Solution:

SOLUTION

You must make the following conversions:

Volume of dilute acid → moles of H?PO? → mass of H?PO? → mass of 85 % H?PO?

0.6000 L dil. acid × $$\displaystyle\frac{{\text{2.80 mol H}_{{3}}\text{PO}_{{4}}}}{\text{1 L dil. acid}}\times\frac{{\text{98.00 g H}_{{3}}\text{PO}_{{4}}}}{{\text{1 mol H}_{{3}}\text{PO}_{{4}}}}$$

$$\displaystyle\frac{{\text{100 g 85 \% H}_{{3}}\text{PO}_{{4}}}}{{\text{85 g H}_{{3}}\text{PO}_{{4}}}}=\text{194 g of 85 \% H}_{{3}}\text{PO}_{{4}}$$

Option 4

3 Hydrogen

5.0 cm³ of H₂O₂ liberates 0.508 g of iodine from an acidified KI solution. The strength of H₂ O₂ solution in terms of volume strength at STP is:

1) 6.48 volumes
2) 4.48 volumes
3) 7.68 volumes
4) none

Option 1

Option 2 Solution:

Solution

Moles of iodine= $$\cfrac {0.508}{254}$$ moles

Now, $$M_1V_1=M_2V_2$$

$$M_1\longrightarrow$$ Molarity of $$H_2O_2$$ solution

$$\therefore$$ $$V_1\longrightarrow$$ Volume of $$H_2O_2$$

$$M_2\longrightarrow$$ Molarity of iodine

$$V_2\longrightarrow$$ $$1000$$ $$ml$$

$$\therefore$$ $$M_1\times 5=\cfrac {0.508}{254}\times 1000$$

$$\therefore$$ $$M_1=0.4$$ $$M$$

The reaction can be given as,

$$H_2O_2+2I^{-}+2H^{+}\longrightarrow2H_2O+I_2$$

$$\therefore$$ n-factor for $$H_2O_2=2$$

$$\therefore$$ Normality of $$H_2O_2=2\times 0.4=0.8N$$

$$\therefore$$ Volume strength of $$H_2O_2=0.8\times 5.6$$

=$$4.48$$ volumes

Option 3

Option 4

4 Haloalkanes and Haloarenes

The compound if contains an even number 'n' of chiral carbons, but the molecule can be divided into two equal and si,ilar halves , then how many optical active form it will have :

1) 2ⁿ
2) 2⁽ⁿ⁻¹⁾
3) 2^(n−1)/2
4) 2⁽ⁿ⁻¹⁾−2^(n−1)/2

Option 1

Option 2 Solution:

SOLUTION

If compound containan even nuumber 'n' of chiral carbons, but the molecule can be divided into two equal and similar halves, then 2ⁿ⁻¹ is the number of optically active forms.

Option 3

Option 4

5 P � Block Elements

Which atom contains exactly 15 protons?

1. P-32
2. S-32
3. O-15
4. N-15

Option 1 Solution:

An atom with 15 protons also has an atomic number of 15. Refer to the Periodic Table for element 15 (phosphorus).

Option 2

Option 3

Option 4

6 Classification of Elements and Periodicity

The first ionization potential of four consecutive elements in the second period of the periodic table are 8.3, 11.3, 14.5 and 13.6 eV respectively. Which one of the following is the first ionization potential of carbon?

1) 13.6
2) 11.3
3) 8.3
4) 14.5

Option 1

Option 2 Solution:

SOLUTION

The first ionization potential of carbon is 11.3 V
The four consecutive elements are B,C,N and O. We know that the first ionization potential of N is higher than the first ionization potential of O as in case of N, the electron is to be removed from stable half-filled 2p orbital.

Option 3

Option 4

7 Solid State

Diamagnetism is the property of

1) non-transition metals

2) completely filled electronic subshells

3) unpaired electrons

4) nucleons

Option 1

Option 2 Solution:

Option 3

Option 4

8 Solid State

A solid X⁺Y^- has a bcc structure. If the distance of closest approach between the ions is 173 pm, the edge length of the cell is

1) √2 pm
2) (√3/2) pm
3) 200 pm
4) 400 pm

Option 1

Option 2

Option 3 Solution:

SOLUTION

In bcc
distance closest approach = a(√3/2)
173 = a(√3/2) a ≈200 pm

Option 4

9 Titrations

25 mL of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave titer value of 35 mL. The molarity of barium hydroxide solution was :

1) 0.07

2) 0.14

3) 0.28

4) 0.35

Option 1 Solution:

Solution

The reaction involved in the titration is:

Ba(OH)₂ + 2HCl -------> BaCl₂ + 2H₂O

The equation that is used to find the molarity of unknown solutions in titrations is:

M₁V₁/n₁ = M₂V₂/n₂

Where

M₁ and M₂ are, respectively, the molarities of Ba(OH)₂ & HCl solutions.

V₁ and V₂ are, respectively, the volumes of Ba(OH)₂ & HCl solutions.

n₁ = Stoichiometric coefficient of Ba(OH)₂ = 1

n₂ = Stoichiometric coefficient of HCl = 2

Therefore:

molarity of barium hydroxide solution, M₁ = (M₂V₂/n₂) x (n₁/V₁) = (0.1 x 35 / 2) x (1/25) = 0.07 M

Option 2

Option 3

Option 4

10 S � Block Elements

The pair of amphoteric hydroxides is:

1) Al(OH)₃and LiOH
2) B(OH)₃ and Be(OH)₂
3) Be(OH)₂ and Zn(OH)₂
4) Be(OH)₂and Mg(OH)₂

Option 1

Option 2

Option 3 Solution:

Solution

Beryllium hydroxide and zinc hydroxide are amphoteric in nature because they react with both, acid and base.
Zinc hydroxide as acid and base:

Zn(OH)₂ + 4H⁺ → Zn²⁺+ 2H₂O

Zn(OH)₂ + 2OH⁻ → [Zn(OH)₄]²⁻

Beryllium hydroxide as acid and base:

Be(OH)₂+ 2HCl → BeCl₂+2H₂O

Be(OH)₂+ 2NaOH → Na₂[Be(OH)₄]

Option 4

NSEC CHEMISTRY

DAILY CHEMISTRY TEST QUESTIONS

Question Bank

Solution

SOLUTION

If compound containan even nuumber 'n' of chiral carbons, but the molecule can be divided into two equal and similar halves, then 2ⁿ⁻¹ is the number of optically active forms.

Solution

Solution

MOCK TEST

NSEC CHEMISTRY

DAILY CHEMISTRY TEST QUESTIONS

Question Bank

Solution

SOLUTION

If compound containan even nuumber 'n' of chiral carbons, but the molecule can be divided into two equal and similar halves, then 2n−1 is the number of optically active forms.

Solution

Solution

MOCK TEST

If compound containan even nuumber 'n' of chiral carbons, but the molecule can be divided into two equal and similar halves, then 2ⁿ⁻¹ is the number of optically active forms.