NEET Chemistry Some Basic Concepts Of Chemistry Questions Solved

#11 | Questions on Equivalent Concept

#12 | Normality of Mixed Solution

Concept Questions :-

Concentration Based Problem

elem

(a) Given, molarity of solution=2

Volume of solution = 250mL=250/1000=1/4L

Molar mass of HNO₃ = 1+14+3x16=63g mol^-1

Molarity = $\frac{w e i g h t o f H N O_{3}}{M o l e c u l a r m a s s o f H N O_{3} \times v o l u m e o f s o l u t i o n (L)}$

Weight of HNO₃ = molarity x molecular massx volume(L) = 2x63x1/4g=31.5g

It is the weight of 100% HNO₃

But the given acid is 70% HNO₃

Its weight = 31.5x100/70g=45g

Difficulty Level:

(a)
52%
(b)
24%
(c)
18%
(d)
8%

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Mole fraction of the solute in a 1.00 molal aqueous solution is

(a) 0.0177

(b) 0.0344

(d) 0.1770

Q2:

Concept Videos :-

#1 | Mole Concept

#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

elem

(a) 1.00 molal aqueous solution = 1.0 mole in 1000g water

n_solute = 1; W_solvent =1000g

n_solvent = 1000/18 = 55.56

X_solute = $\frac{n_{s o l u t e}}{n_{s o l u t e} + n_{s o l v e n t}}$

X_solute= 1/1+55.56 = 0.0177

Difficulty Level:

(a)
55%
(b)
22%
(c)
14%
(d)
11%

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6.02x10²⁰ molecules of urea are present in 100mL of its solution. The concentration of solution is

(a) 0.02M

(b) 0.01M

(d) 0.1M

Q3:

Concept Videos :-

#1 | Mole Concept

#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

elem

(b) Given, number of molecules of urea= 6.02x10²⁰

Number of moles = 6.02x10²⁰ /N_A

=6.02x10²⁰ /6.02x10²³ =1x10^-3 mol

Volume of the solution = 100mL = 100/1000L =0.1L

Concentration of urea solution (in mol L^-1)

=1x10^-3 /0.1 mol L^-1

=1x10^-2 mol L^-1

=0.01 mol L^-1

Difficulty Level:

(a)
14%
(b)
60%
(c)
17%
(d)
11%

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Which has maximum number of molecules?

(a) 7g N₂

(b) 2g H₂

(d) 16g O₂

Q4:

Concept Videos :-

#1 | Mole Concept

#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

elem

(b) In 7g nitrogen, number of molecules = $\frac{7.0}{28}$ mol = 0.25 x N_A molecules

where, N_A = Avogadro number = 6.023 x 10²³

In 2g of H₂= 2/2 mol =1x N_A molecules

In 16 g of NO₂= 16/46 mol = 0.348 x N_A molecules

In 16 g of O₂ = 16/32 mol = 0.5 x N_A molecules

Hence, maximum number of molecules are present in 2g of H₂.

Difficulty Level:

(a)
8%
(b)
63%
(c)
24%
(d)
8%

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The number of moles of KMnO₄ reduced by one mole of KI in alkaline medium is

(a) one fifth

(b) five

(d) two

Q5:

Concept Videos :-

#5 | Stoichiometry

#6 | Questions on POAC

#13 | Calculation of % composition & % Purity

Concept Questions :-

Equation Based Problem

elem

(d) as per the balanced equation, one mole of KI in basic medium need 2 moles of KMnO₄

Video Solution:

Difficulty Level:

(a)
25%
(b)
24%
(c)
29%
(d)
24%

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The number of moles of oxygen in 1L of air containing 21% oxygen by volume, under standard conditions, is

(a) 0.0093 mole

(b) 2.10 moles

(d) 0.21 mole

Q6:

Concept Videos :-

#1 | Mole Concept

#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

elem

(a) Volume oxygen in 1L of air = 21/100x1000 = 210mL

22400 mL volume at STP is occupied by oxygen = 1mole

Therefore, number of moles occupied by 210mL = 210/22400 = 0.0093mol

Difficulty Level:

(a)
53%
(b)
18%
(c)
15%
(d)
15%

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What volume of oxygen gas (O₂) measured at $0 ° C$ and 1 atm, is needed to burn completely 1L of propane gas (C₃H₈) measured under the same conditions?

(a) 7L

(b) 6L

(d) 10L

Q7:

Concept Videos :-

#5 | Stoichiometry

#6 | Questions on POAC

#13 | Calculation of % composition & % Purity

Concept Questions :-

Equation Based Problem

elem

22.4L 5x22.4L

For the combustion of 22.4L propane, oxygen required = 5 x 22.4L

For the combustion of 1L of propane oxygen required = 5x22.4/22.4L

=5L

Difficulty Level:

(a)
12%
(b)
17%
(c)
64%
(d)
8%

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1 g of magnesium is burnt with 0.56g of oxygen in a closed vessel. Which reactant is left in excess and how much?

(At. weight of Mg = 24, O=16)

(a) Mg, 0.16g

(b) O₂, 0.16g

(d) O₂, 0.28g

Q8:

Concept Videos :-

#5 | Stoichiometry

#6 | Questions on POAC

#13 | Calculation of % composition & % Purity

Concept Questions :-

Equation Based Problem

elem

(a) The balanced chemical equation is

From the above equation, it is clear that, 24g of Mg reacts with 16g of O₂ .

Thus,1.0 g of Mg reacts with 16/24g of O₂ = 0.67g of O₂

But only 0.56g of O₂ is available which is less than 0.67g. Thus,O₂ is the limiting reagent. Further, 16g of O₂ reacts with 24g of Mg.

0.56g of O₂ will react with Mg = 24/16x0.56 = 0.84g

Amount of Mg left unreacted = (1.0-0.84)g Mg = 0.16g of Mg

Difficulty Level:

(a)
50%
(b)
20%
(c)
25%
(d)
8%

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