Solution & Explanation
### Related Formula
mathrmC_mathrmxmathrmH_mathrmy(g) + left(mathrmx + fracmathrmy4right)mathrmO_2(g) longrightarrow mathrmxCO_2(g) + fracmathrmy2mathrmH_2mathrmO_(ell)$$\mathrm{C}_\mathrm{x}\mathrm{H}_\mathrm{y(g)} + \left(\mathrm{x} + \frac{\mathrm{y}}{4}\right)\mathrm{O}_{2(g)} \longrightarrow \mathrm{xCO}_{2(g)} + \frac{\mathrm{y}}{2}\mathrm{H}_2\mathrm{O}_{(\ell)}$$
### Core Logic
Let the volume of hydrocarbon be V = 80$V = 80$ mL.
Initial volume of O_2 = 264$O_2 = 264$ mL.
At 273 K, H_2O$H_2O$ is liquid, so its volume is neglected.
Volume of CO_2$CO_2$ formed = 80x$80x$ mL.
Volume of O_2$O_2$ used = 80left(x + fracy4right)$80\left(x + \frac{y}{4}\right)$ mL.
Unreacted O_2$O_2$ = 264 - 80left(x + fracy4right)$264 - 80\left(x + \frac{y}{4}\right)$ mL.
Total residual volume = V_CO_2 + V_unreacted \ O_2 = 224$V_{CO_2} + V_{unreacted \ O_2} = 224$ mL.
80x + 264 - 80left(x + fracy4right) = 224$$80x + 264 - 80\left(x + \frac{y}{4}\right) = 224$$
264 - frac80y4 = 224$$264 - \frac{80y}{4} = 224$$
40 = 20y implies y = 2$$40 = 20y \implies y = 2$$
After treatment with KOH, CO_2$CO_2$ is absorbed. The remaining volume is unreacted O_2$O_2$, which is 64 mL.
264 - 80left(x + fracy4right) = 64$$264 - 80\left(x + \frac{y}{4}\right) = 64$$
Substitute y = 2$y = 2$:
264 - 80left(x + frac12right) = 64$$264 - 80\left(x + \frac{1}{2}\right) = 64$$
264 - 80x - 40 = 64$$264 - 80x - 40 = 64$$
224 - 80x = 64$224 - 80x = 64$
80x = 160 implies x = 2$$80x = 160 \implies x = 2$$
The hydrocarbon is mathrmC_2mathrmH_2$\mathrm{C}_2\mathrm{H}_2$.
### Pattern Recognition
Volume decrease by KOH indicates the volume of CO_2$CO_2$ produced.
V_CO_2 = 224 - 64 = 160$V_{CO_2} = 224 - 64 = 160$ mL.
V_HC = 80$V_{HC} = 80$ mL. So x = frac16080 = 2$x = \frac{160}{80} = 2$.
Total volume reduction = 264 - 64 = 200 mL (O_2$O_2$ consumed).
O_2$O_2$ consumed = 80(x + y/4) = 200 implies 2 + y/4 = 2.5 implies y/4 = 0.5 implies y = 2$80(x + y/4) = 200 \implies 2 + y/4 = 2.5 \implies y/4 = 0.5 \implies y = 2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
More Some Basic Concepts of Chemistry Previous-Year Questions — Page 2
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 mathrmN_2 = 28text g mol^-1$\mathrm{N}_2 = 28\text{ g mol}^{-1}$, mathrmO_2 = 32text g mol^-1$\mathrm{O}_2 = 32\text{ g mol}^{-1}$ and mathrmAr = 40text g mol^-1$\mathrm{Ar} = 40\text{ g mol}^{-1}$ respectively)
- A. 4.26, 19.3
- B. 2.59, 11.85
- C. 5.46, 17.8
- D. 2.96, 11.2
Solution
### Related Formula
P_i = X_i cdot P_texttotal = fracn_in_texttotal cdot P_texttotal$$P_i = X_i \cdot P_{\text{total}} = \frac{n_i}{n_{\text{total}}} \cdot P_{\text{total}}$$
Ratio of partial pressures:
fracP_AP_B = fracn_An_B$$\frac{P_A}{P_B} = \frac{n_A}{n_B}$$
### Core Logic
Assume a sample of dry air with total mass = 100 text g$= 100 \text{ g}$:
- Mass of mathrmN_2 = 70.0 text g$\mathrm{N}_2 = 70.0 \text{ g}$
- Mass of mathrmO_2 = 27.0 text g$\mathrm{O}_2 = 27.0 \text{ g}$
- Mass of mathrmAr = 3.0 text g$\mathrm{Ar} = 3.0 \text{ g}$
Now, convert masses to moles:
n_mathrmN_2 = frac70.028 = 2.5 text moles$$n_{\mathrm{N}_2} = \frac{70.0}{28} = 2.5 \text{ moles}$$
n_mathrmO_2 = frac27.032 = 0.84375 text moles$$n_{\mathrm{O}_2} = \frac{27.0}{32} = 0.84375 \text{ moles}$$
n_mathrmAr = frac3.040 = 0.075 text moles$$n_{\mathrm{Ar}} = \frac{3.0}{40} = 0.075 \text{ moles}$$
Calculate ratios:
(i) Ratio of partial pressure of nitrogen to oxygen:
fracP_mathrmN_2P_mathrmO_2 = fracn_mathrmN_2n_mathrmO_2 = frac2.50.84375 approx 2.96$$\frac{P_{\mathrm{N}_2}}{P_{\mathrm{O}_2}} = \frac{n_{\mathrm{N}_2}}{n_{\mathrm{O}_2}} = \frac{2.5}{0.84375} \approx 2.96$$
(ii) Ratio of partial pressure of oxygen to argon:
fracP_mathrmO_2P_mathrmAr = fracn_mathrmO_2n_mathrmAr = frac0.843750.075 approx 11.25 approx 11.2$$\frac{P_{\mathrm{O}_2}}{P_{\mathrm{Ar}}} = \frac{n_{\mathrm{O}_2}}{n_{\mathrm{Ar}}} = \frac{0.84375}{0.075} \approx 11.25 \approx 11.2$$
### 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.
### Evaluation Rubric / Model Answer
null
### 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
A 20 text mL$20 \text{ mL}$ sample of a sodium iodide solution yields 4.74 text g$4.74 \text{ g}$ 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:
textNa = 23, \, textI = 127, \, textAg = 108, \, textN = 14, \, textO = 16 text g mol^-1$\text{Na} = 23, \, \text{I} = 127, \, \text{Ag} = 108, \, \text{N} = 14, \, \text{O} = 16 \text{ g mol}^{-1}$.
Numerical Answer. Answer: 1 to 1
Solution
### Related Formula
Precipitation reaction stoichiometry:
textNaI(aq) + textAgNO_3text(aq) longrightarrow textAgI(s) + textNaNO_3text(aq)$$\text{NaI(aq)} + \text{AgNO}_3\text{(aq)} \longrightarrow \text{AgI(s)} + \text{NaNO}_3\text{(aq)}$$
Molarity calculation formula:
M = fractextMoles of solute (NaI)textVolume of solution in Liters (L)$$M = \frac{\text{Moles of solute (NaI)}}{\text{Volume of solution in Liters (L)}}$$
### Execution
Step 1: Determine the molar mass of the Silver Iodide (textAgI$\text{AgI}$) precipitate:
textMolar Mass of AgI = 108 + 127 = 235 text g mol^-1$$\text{Molar Mass of AgI} = 108 + 127 = 235 \text{ g mol}^{-1}$$
Step 2: Calculate the moles of textAgI$\text{AgI}$ precipitated:
textMoles of AgI = frac4.74 text g235 text g mol^-1 approx 0.02017 text mol$$\text{Moles of AgI} = \frac{4.74 \text{ g}}{235 \text{ g mol}^{-1}} \approx 0.02017 \text{ mol}$$
Step 3: Apply the 1:1 reaction stoichiometry to find the moles of textNaI$\text{NaI}$:
textMoles of NaI = textMoles of AgI = 0.02017 text mol$$\text{Moles of NaI} = \text{Moles of AgI} = 0.02017 \text{ mol}$$
Step 4: Compute the molarity of the solution, converting 20 text mL$20 \text{ mL}$ to 0.020 text L$0.020 \text{ L}$:
textMolarity [NaI] = frac0.02017 text mol0.020 text L = 1.0085 text M$$\text{Molarity [NaI]} = \frac{0.02017 \text{ mol}}{0.020 \text{ L}} = 1.0085 \text{ M}$$
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.
### Evaluation Rubric / Model Answer
null
### 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.
mathrmC:14.5\% quad mathrmCl:64.46\% quad mathrmH:1.8\%$$\mathrm{C}:14.5\% \quad \mathrm{Cl}:64.46\% \quad \mathrm{H}:1.8\%$$
The empirical formula mass of the compound (X) is mathrmx times 10^-1$\mathrm{x} \times 10^{-1}$. The value of mathrmx$\mathrm{x}$ is:
(Given molar mass in mathrmg\,mol^-1$\mathrm{g\,mol^{-1}}$ 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:
\%mathrmO = 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24\%$$\%\mathrm{O} = 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24\%$$
### Step 2: Calculate Molar Ratios
Divide each mass percentage by its respective atomic weight:
- mathrmC: frac14.512 = 1.208$\mathrm{C}: \frac{14.5}{12} = 1.208$
- mathrmCl: frac64.4635.5 = 1.815$\mathrm{Cl}: \frac{64.46}{35.5} = 1.815$
- mathrmH: frac1.81 = 1.800$\mathrm{H}: \frac{1.8}{1} = 1.800$
- mathrmO: frac19.2416 = 1.202$\mathrm{O}: \frac{19.24}{16} = 1.202$
### Step 3: Find Simple Integer Ratio
Divide by the lowest ratio value (1.202$1.202$):
- mathrmC: frac1.2081.202 approx 1 rightarrow times 2 = 2$\mathrm{C}: \frac{1.208}{1.202} \approx 1 \rightarrow \times 2 = 2$
- mathrmCl: frac1.8151.202 approx 1.5 rightarrow times 2 = 3$\mathrm{Cl}: \frac{1.815}{1.202} \approx 1.5 \rightarrow \times 2 = 3$
- mathrmH: frac1.8001.202 approx 1.5 rightarrow times 2 = 3$\mathrm{H}: \frac{1.800}{1.202} \approx 1.5 \rightarrow \times 2 = 3$
- mathrmO: frac1.2021.202 = 1 rightarrow times 2 = 2$\mathrm{O}: \frac{1.202}{1.202} = 1 \rightarrow \times 2 = 2$
Thus, the empirical formula is mathrmC_2mathrmH_3mathrmCl_3mathrmO_2$\mathrm{C}_2\mathrm{H}_3\mathrm{Cl}_3\mathrm{O}_2$.
### Step 4: Compute Mass
Empirical formula mass calculation:
textMass = (2 times 12) + (3 times 1) + (3 times 35.5) + (2 times 16)$$\text{Mass} = (2 \times 12) + (3 \times 1) + (3 \times 35.5) + (2 \times 16)$$
textMass = 24 + 3 + 106.5 + 32 = 165.5\,mathrmg\,mol^-1$$\text{Mass} = 24 + 3 + 106.5 + 32 = 165.5\,\mathrm{g\,mol}^{-1}$$
Expressing in the requested format:
165.5 = 1655 times 10^-1 Rightarrow x = 1655$$165.5 = 1655 \times 10^{-1} \Rightarrow x = 1655$$
### Pattern Recognition
Sees: Multi-element empirical calculation.
Trap: Forgetting to compute Oxygen by missing that the percentages do not sum to 100% initial value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q48
jee_main_2025_28_jan_morning
Molarity of Solutions
The molarity of a 70\%$70\%$ (mass/mass) aqueous solution of a monobasic acid (X) is \_\_\_\_\_mathrmM$\_\_\_\_\_\mathrm{M}$ (Nearest integer)
[Given : Density of aqueous solution of (X) is 1.25mathrmg\,mathrmmL^-1$1.25\mathrm{g}\,\mathrm{mL}^{-1}$
Molar mass of the acid is 70mathrmg\,mol^-1$70\mathrm{g\,mol}^{-1}$]
Numerical Answer. Answer: 12.5 to 13.5
Solution
### Related Formula
Molarity formula based on mass percentage (w/w$w/w$) and density (d$d$):
textMolarity = frac\%(w/w) times d times 10textMolar Mass of solute$$\text{Molarity} = \frac{\%(w/w) \times d \times 10}{\text{Molar Mass of solute}}$$
### Step 1: Substitute Values
Given values: \% = 70$\% = 70$, d = 1.25\,mathrmg\,mL^-1$d = 1.25\,\mathrm{g\,mL}^{-1}$, textMolar Mass = 70\,mathrmg\,mol^-1$\text{Molar Mass} = 70\,\mathrm{g\,mol}^{-1}$.
textMolarity = frac70 times 1.25 times 1070 = 1.25 times 10 = 12.5\,mathrmM$$\text{Molarity} = \frac{70 \times 1.25 \times 10}{70} = 1.25 \times 10 = 12.5\,\mathrm{M}$$
Rounding to the nearest integer gives 13 (or 12.5 as written in standard templates; let us provide 13 matching nearest integer constraints).
### Pattern Recognition
Sees: Conversion of mass percentage to molarity tracking.
Shortcut: Using the classic shortcut formula frac\% times d times 10M$\frac{\% \times d \times 10}{M}$ simplifies the arithmetic immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q33
jee_main_2025_03_april_morning
Mole Concept - Number of Atoms
Among 10^-9text g$10^{-9}\text{ g}$ (each) of the following elements, which one will have the highest number of atoms ?
Element: Pb, Po, Pr and Pt
Solution
### Related Formula
The number of atoms in a given mass of an element is calculated using:
textNumber of atoms = fractextMass (g)textMolar Mass (g/mol) times N_A$$\text{Number of atoms} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \times N_A$$
### Core Logic
Since the mass (10^-9text g$10^{-9}\text{ g}$) is identical for all samples, the number of atoms is inversely proportional to the molar mass of the element:
textNumber of atoms propto frac1textMolar Mass
$$\text{Number of atoms} \propto \frac{1}{\text{Molar Mass}}
$$
### Step 1: Molar Mass Values Comparison
Let us check the molar masses of the listed elements:
* textMolar Mass of Po approx 209text g/mol$\text{Molar Mass of Po} \approx 209\text{ g/mol}$
* textMolar Mass of Pr approx 141text g/mol$\text{Molar Mass of Pr} \approx 141\text{ g/mol}$
* textMolar Mass of Pb approx 207text g/mol$\text{Molar Mass of Pb} \approx 207\text{ g/mol}$
* textMolar Mass of Pt approx 195text g/mol$\text{Molar Mass of Pt} \approx 195\text{ g/mol}$
### Step 2: Conclusion
Praseodymium (Pr) has the least molar mass (141text g/mol$141\text{ g/mol}$), meaning it will yield the maximum total number of atoms for the specified mass sample.
### Pattern Recognition
Shortcut: Equal mass given
ightarrow$
ightarrow$ Lighter atoms mean more atoms per gram. Find the element with the lowest atomic mass value from the choices.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Concepts of Chemistry
