Some Basic Concepts of Chemistry JEE JEE Main| JEE Main Important Questions

Q 1 :

Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is: [2025]

2 m
2.79 m
1.79 m
3 m

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(2)

Molar mass of NaCl = 23 + 35.5 = 58.5 g ${mol}^{- 1}$

3 M NaCl solution has:

3 mol NaCl in 1000 mL solution

$i.e 3 \times 58.5 g NaCl in (1000 mL \times 1.25 {g mL}^{- 1}) solution$

$i . e$ $175.5 g NaCl in 1250 g solution$

$i . e$ $175.5 g NaCl in (1250 - 175.5) g solvent$

$i . e$ $175.5 g NaCl in 1074.5 g solvent$

$i . e$ $3 mol NaCl in 1074.5 g solvent$

Molality (m) : $= \frac{n_{solute}}{W_{solvent} (in g)} \times 1000$

$= \frac{3}{1074.5} \times 1000 = 2.79 m$

Q 2 :

Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______.

Given: Density of nitric acid solution is 1.25 g/mL. [2025]

45
55
32
40

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(3)

$\frac{% w}{w} = \frac{w_{solute}}{w_{solution}} \times 100$

$75 = \frac{30}{w_{solution}} \times 100$

$w_{solution} = 40 g$

$V_{solution} = \frac{w_{solution}}{d_{solution}} = \frac{40 g}{1.25 {g mL}^{- 1}} = 32$ $mL$

Q 3 :

The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is _____ $\times 10^{- 1}$ M (Nearest integer)

[Given: Density of aqueous solution of (X) is 1.25 g ${mL}^{- 1}$ Molar mass of the acid is 70 g ${mol}^{- 1}$ ] [2025]

(125)

$M = \frac{(% w / w) \times d \times 10}{M_{solute}}$

$M = \frac{70 \times 1.25 \times 10}{70} M = 125 \times 10^{- 1} M$

Q 4 :

The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is ______ $\times 10^{- 2}$ mL (Nearest integer) [2023]

(12)

$Molarity of acid = \frac{1.2 \times 10^{3}}{24.2} = \frac{1000}{20} = 50 M$

$Neutralization reaction:$

$HA + NaOH ⟶ NaA + H_{2} O$

$M_{1} V_{1} = M_{2} V_{2}$

$[50] \times V = [0.24 \times 25]$

$V = 0.12 ml$

Q 5 :

The volume of HCl, containing 73 g $L^{- 1}$ , required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water is _____ mL. (Nearest Integer

(Given: Molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g ${mol}^{- 1}$ respectively) [2023]

(15)

$2 Na + 2 H_{2} O ⟶ 2 NaOH + H_{2}$

$Number of moles of Na = Number of moles of NaOH$

$n_{NaOH} = \frac{0.69}{23} = 0.03$

$NaOH + HCl ⟶ NaCl + H_{2} O$

$Moles of HCl = moles of NaOH$

${(Molarity \times V)}_{HCl} = Number of moles of NaOH$

$(\frac{73}{36.5 \times 1}) \times V = 0.03$

$\Rightarrow V = 15 \times 10^{- 3} Lit = 15 mL.$

Q 6 :

For the given reaction;

$C a C O_{3} + 2 H C l ⟶ C a C l_{2} + H_{2} O + C O_{2} $

If 90 g $C a C O_{3}$ is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 1.13 g $m L^{- 1}$ , then which of the following option is correct ?

Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g $m o l^{- 1}$ respectively. [2026]

97.30 g of HCl reacted
64.97 g of HCl remains unreacted
60.32 g of HCl remains unreacted
32.85 g of $C a C O_{3} $ remains unreacted

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(2)

$Density of HCl solution (d) = 1.13 g/mL$

$V = 300 mL$

$Wt. of HCl solution = 339 g$

$Wt. of HCl = 339 \times \frac{38.55}{100} = 130.68 g$

$\begin{matrix} (LR) \\ {CaCO}_{3} & + & 2 HCl & \to & {CaCl}_{2} & + & H_{2} O & + & {CO}_{2} \\ \frac{90}{100} & \frac{130.68}{36.5} \\ = 0.90 mole & = 3.58 mole \end{matrix}$

$Moles of HCl remained = 1.78 mole$

$Mass of HCl remained = 64.97 g$

Q 7 :

29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____

(4)

$\frac{4 m L}{M} = (\frac{% (w / w) \times d \times 10}{molar mass}) = \frac{(29.2) \times 1.25 \times 10}{36.5} = 10 M$

$According to dilution, equation, M_{i} V_{i} = M_{f} V_{f}$

$V_{f} = \frac{0.4 \times 100}{10} = 4 m L$

Q 8 :

Given below are two statements: one labelled as Assertion A and other is labelled as Reason R

Assertion A: 1.24 g of hydrated hypo dissolved in water to make 250.0 ml solution result in 0.2 M hypo solution

Reason R: Molar mass of hydrated hypo is 248 g ${mol}^{- 1}$

In the light of the above statements. Choose the correct answer from the options given below

A is true but R is false
A is false but R is true
Both A and R are true but R is not the correct explanation of A
Both A and R are true and R is the correct explanation of A

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(2)

$N a_{2} S_{2} O_{3} \cdot 5 H_{2} O called hypo$

$Mol.wt = 248$

$M = \frac{1.24}{248} \times \frac{1000}{250} = 0.02$

Topic Question Set

Q 1 :

Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is: [2025]

Q 2 :

Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______.

Given: Density of nitric acid solution is 1.25 g/mL. [2025]

Q 3 :

The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is _____ $\times 10^{- 1}$ M (Nearest integer)

[Given: Density of aqueous solution of (X) is 1.25 g ${mL}^{- 1}$ Molar mass of the acid is 70 g ${mol}^{- 1}$ ] [2025]

Q 4 :

The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is ______ $\times 10^{- 2}$ mL (Nearest integer) [2023]

Q 5 :

The volume of HCl, containing 73 g $L^{- 1}$ , required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water is _____ mL. (Nearest Integer

(Given: Molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g ${mol}^{- 1}$ respectively) [2023]

Q 7 :

29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____

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Topic Question Set

Q 1 : Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is: [2025]

Q 2 : Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______. Given: Density of nitric acid solution is 1.25 g/mL. [2025]

Q 3 : The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is _____ ×10-1 M (Nearest integer) [Given: Density of aqueous solution of (X) is 1.25 g mL-1 Molar mass of the acid is 70 g mol-1] [2025]

Q 4 : The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is ______ ×10-2 mL (Nearest integer) [2023]

Q 5 : The volume of HCl, containing 73 g L-1, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water is _____ mL. (Nearest Integer (Given: Molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g mol-1 respectively) [2023]

Q 7 : 29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____

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Q 1 :

Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is: [2025]

Q 2 :

Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______.

Given: Density of nitric acid solution is 1.25 g/mL. [2025]

Q 3 :

The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is _____ $\times 10^{- 1}$ M (Nearest integer)

[Given: Density of aqueous solution of (X) is 1.25 g ${mL}^{- 1}$ Molar mass of the acid is 70 g ${mol}^{- 1}$ ] [2025]

Q 4 :

The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is ______ $\times 10^{- 2}$ mL (Nearest integer) [2023]

Q 5 :

The volume of HCl, containing 73 g $L^{- 1}$ , required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water is _____ mL. (Nearest Integer

(Given: Molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g ${mol}^{- 1}$ respectively) [2023]

Q 7 :

29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____