The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is ___

Year	2026	2025	2024	Total
Questions	8	15	7	30

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 3

Q37 jee_main_2025_03_april_evening Stoichiometry of Gas Evolution

Mass of magnesium required to produce

220~mL

of hydrogen gas at STP on reaction with excess of dil. HCl is : Given: Molar mass of Mg is

24~g~mol⁻¹

A. 235.7 g
B. 0.24 mg
C. 236 mg
D. 2.444 g

Solution

Related Formula

The balanced chemical equation for the displacement reaction is:

Mg(s) + 2HCl(aq) arrow MgCl₂(aq) + H₂(g)

At STP, 1 mole of any ideal gas occupies a volume of $22.4~L = 22400~mL$ $22.4\mathrm{~L} = 22400\mathrm{~mL}$.

Core Logic

From the stoichiometry of the reaction:

1 mole of $Mg$ $\mathrm{Mg}$ ( $24~g$ $24\mathrm{~g}$) produces 1 mole of $H₂$ $\mathrm{H}_2$ ( $22400~mL$ $22400\mathrm{~mL}$ at STP).

Step 1: Calculate moles of $H₂$ $\mathrm{H}_2$ gas produced

nH₂ = 220~mL22400~mL/mol \approx 9.8214 \times 10⁻³~mol

Step 2: Calculate mass of Magnesium required Since the molar ratio of$ Step 2: Calculate mass of Magnesium required Since the molar ratio of $\mathrm{Mg}to$ to $\mathrm{H}_2is$ is $1:1:

nMg = 9.8214 \times 10⁻³~mol

Mass of Mg = 9.8214 \times 10⁻³~mol \times 24~g/mol

Mass of Mg \approx 0.2357~g = 235.7~mg \approx 236~mg

This matches Option (3). Pattern Recognition Always keep a close eye on unit prefixes in options. A mass of$ This matches Option (3). Pattern Recognition Always keep a close eye on unit prefixes in options. A mass of $0.2357\mathrm{~g}corresponds to$ corresponds to $235.7\mathrm{~mg}, which rounds directly to$, which rounds directly to $236\mathrm{~mg}, whereas$, whereas $235.7\mathrm{~g}$ is off by a factor of 1000. Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

Q34 jee_main_2025_07_april_morning Dalton's Law of Partial Pressure

At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm, then calculate the ratio of followings respectively : (i) partial pressure of nitrogen gas to partial pressure of oxygen gas (ii) partial pressure of oxygen gas to partial pressure of argon gas (Given: Molar mass of

N₂ = 28 g mol⁻¹

O₂ = 32 g mol⁻¹

and

Ar = 40 g mol⁻¹

respectively)

A. 4.26, 19.3
B. 2.59, 11.85
C. 5.46, 17.8
D. 2.96, 11.2

Solution

Related Formula

Pᵢ = Xᵢ \cdot Ptotal = nᵢntotal \cdot Ptotal

Ratio of partial pressures:

(PA)/(PB) = (nA)/(nB)

Core Logic

Assume a sample of dry air with total mass $= 100 g$ $= 100 \text{ g}$:

Mass of $N₂ = 70.0 g$ $\mathrm{N}_2 = 70.0 \text{ g}$
Mass of $O₂ = 27.0 g$ $\mathrm{O}_2 = 27.0 \text{ g}$
Mass of $Ar = 3.0 g$ $\mathrm{Ar} = 3.0 \text{ g}$

Now, convert masses to moles:

nN₂ = (70.0)/(28) = 2.5 moles

nO₂ = (27.0)/(32) = 0.84375 moles

nAr = (3.0)/(40) = 0.075 moles

Calculate ratios: (i) Ratio of partial pressure of nitrogen to oxygen:

PN₂PO₂ = nN₂nO₂ = (2.5)/(0.84375) \approx 2.96

(ii) Ratio of partial pressure of oxygen to argon:

PO₂PAr = nO₂nAr = (0.84375)/(0.075) \approx 11.25 \approx 11.2

Pattern Recognition

Since total pressure cancels out in a ratio of partial pressures, we only need to calculate the mole ratio directly from the given mass percentages divided by their respective molar masses.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry Class 11 Physics: Kinetic Theory of Gases

Q48 jee_main_2025_08_april_evening Stoichiometry and Molarity

20 mL

sample of a sodium iodide solution yields

4.74 g

of silver iodide precipitate when treated with an excess of silver nitrate solution. The molarity of the initial sodium iodide solution is _________ M (as the nearest integer value). Given molar masses:

Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol⁻¹

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

Precipitation reaction stoichiometry:

NaI(aq) + AgNO₃(aq) AgI(s) + NaNO₃(aq)

Molarity calculation formula:

M = Moles of solute (NaI)Volume of solution in Liters (L)

Execution

Step 1: Determine the molar mass of the Silver Iodide ( $AgI$ $\text{AgI}$) precipitate:

Molar Mass of AgI = 108 + 127 = 235 g mol⁻¹

Step 2: Calculate the moles of $AgI$ $\text{AgI}$ precipitated:

Moles of AgI = 4.74 g235 g mol⁻¹ \approx 0.02017 mol

Step 3: Apply the 1:1 reaction stoichiometry to find the moles of $NaI$ $\text{NaI}$:

Moles of NaI = Moles of AgI = 0.02017 mol

Step 4: Compute the molarity of the solution, converting $20 mL$ $20 \text{ mL}$ to $0.020 L$ $0.020 \text{ L}$:

Molarity [NaI] = 0.02017 mol0.020 L = 1.0085 M

Rounding to the nearest integer value gives 1.

Pattern Recognition

Precipitation reactions involving silver halides follow a strict 1:1 mole ratio between the halide source and the silver precipitate. Converting mass into moles and dividing by the volume in liters quickly yields the molarity.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Q47 jee_main_2025_28_jan_morning Empirical Formula Calculation

Quantitative analysis of an organic compound (X) shows following % composition.

C:14.5% Cl:64.46% H:1.8%

The empirical formula mass of the compound (X) is

x \times 10⁻¹

. The value of

x

is: (Given molar mass in

g mol⁻¹

of C: 12, H: 1, O: 16, Cl: 35.5)

Numerical Answer. Answer: 1655 to 1655

Solution

Step 1: Determine Oxygen Percentage

The total percentage must equal 100%. The remaining composition corresponds to Oxygen:

%O = 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24%

Step 2: Calculate Molar Ratios

Divide each mass percentage by its respective atomic weight:

$C: (14.5)/(12) = 1.208$ $\mathrm{C}: \frac{14.5}{12} = 1.208$
$Cl: (64.46)/(35.5) = 1.815$
$H: (1.8)/(1) = 1.800$ $\mathrm{H}: \frac{1.8}{1} = 1.800$
$O: (19.24)/(16) = 1.202$ $\mathrm{O}: \frac{19.24}{16} = 1.202$

Step 3: Find Simple Integer Ratio

Divide by the lowest ratio value ( $1.202$ $1.202$):

$C: (1.208)/(1.202) \approx 1 arrow \times 2 = 2$
$Cl: (1.815)/(1.202) \approx 1.5 arrow \times 2 = 3$
$H: (1.800)/(1.202) \approx 1.5 arrow \times 2 = 3$
$O: (1.202)/(1.202) = 1 arrow \times 2 = 2$

Thus, the empirical formula is $C₂H₃Cl₃O₂$ .

Step 4: Compute Mass

Empirical formula mass calculation:

Mass = (2 \times 12) + (3 \times 1) + (3 \times 35.5) + (2 \times 16)

Mass = 24 + 3 + 106.5 + 32 = 165.5 g mol⁻¹

Expressing in the requested format:

165.5 = 1655 \times 10⁻¹ \Rightarrow x = 1655

Pattern Recognition

Sees: Multi-element empirical calculation. Trap: Forgetting to compute Oxygen by missing that the percentages do not sum to 100% initial value.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

The molarity of a 70%$70\%$ (mass/mass) aqueous solution of a monobasic acid (X) is ____M$\_\_\_\_\_\mathrm{M}$ (Nearest integer) [Given : Density of aqueous solution of (X) is 1.25g mL⁻¹$1.25\mathrm{g}\,\mathrm{mL}^{-1}$ Molar mass of the acid is 70g mol⁻¹$70\mathrm{g\,mol}^{-1}$]

Solution & Explanation

Related Formula

Step 1: Substitute Values

Pattern Recognition

Chapter Mix

Reference Study Guides

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate moles of H₂$\mathrm{H}_2$ gas produced

Step 2: Calculate mass of Magnesium required

Step 2: Calculate mass of Magnesium required

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Execution

Pattern Recognition

Chapter Mix

Solution

Step 1: Determine Oxygen Percentage

Step 2: Calculate Molar Ratios

Step 3: Find Simple Integer Ratio

Step 4: Compute Mass

Pattern Recognition

Chapter Mix

More Some Basic Concepts of Chemistry Questions — jee_main_2025_28_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Step 1: Calculate moles of $H₂$ $\mathrm{H}_2$ gas produced