Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
The mass of iron converted into Fe$_3$O$_4$ by the action of 18 g of steam is :
(Given : Molar mass of H, O and Fe are 1, 16 and 56 g mol$^{-1}$ respectively)
Assume iron is present in excess :
The ratio of mass percentage (w/w) of C : H in a hydrocarbon is 12 : 1. It has two carbon atoms. The weight (in g) of $CO_2(g)$ formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g mol$^{-1}$ C : 12, H : 1, , O : 16)
(1) 5.68
(2) 11.44
(3) 22.74
(4) 17.05
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Answer: Option (2)
Key idea: Stoichiometry and combustion analysis of hydrocarbons
Given mass ratio C:H = 12:1 and 2 carbon atoms, find molecular formula: Mass of C = $2 \times 12 = 24$ g, Mass of H = $24/12 = 2$ g, so H atoms = 2. Formula is $C_2H_2$
Calculate molar mass of $C_2H_2$: $M = 2(12) + 2(1) = 26$ g/mol
Write combustion equation: $C_2H_2 + \frac{5}{2}O_2 \rightarrow 2CO_2 + H_2O$. Each mole of $C_2H_2$ produces 2 moles of $CO_2$
Calculate moles of hydrocarbon: $n_{C_2H_2} = \frac{3.38}{26} = 0.13$ mol
Complete combustion of X g of an organic compound gave 0.25 g of $CO_2$ and 0.12 g of $H_2O$. If the % of carbon is 25% and of hydrogen is 4.89%, then $X = $ ______ $\times 10^{-3}$ g. (Nearest integer) (Molar mass of C, H and O are 12, 1 and 16 g mol$^{-1}$ respectively.)
(1) 273
(2) 27
(3) 2730
(4) 227
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Answer: Option (1)
Key idea: Stoichiometry and Percentage Composition
Calculate mass of carbon from $CO_2$: Mass of C $= \frac{12}{44} \times 0.25 = 0.0682$ g
Calculate mass of hydrogen from $H_2O$: Mass of H $= \frac{2}{18} \times 0.12 = 0.0133$ g
Use percentage of carbon: If 25% of X = 0.0682 g, then $X = \frac{0.0682}{0.25} = 0.273$ g
Verify with hydrogen: If 4.89% of X = 0.0133 g, then $X = \frac{0.0133}{0.0489} = 0.273$ g
Convert to required format: $X = 0.273$ g $= 273 \times 10^{-3}$ g
In sulphur estimation, $2.0 \times 10^{-3}$ mol of an organic compound (X) (molar mass $76$ g mol$^{-1}$) gave $0.4813$ g of barium sulphate (molar mass $233$ g mol$^{-1}$). The percentage of sulphur in the compound (X) is __________ $\times 10^{-1}$ % (Nearest integer)
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Answer: 435
Key idea: Carius Method for Sulphur Estimation - Stoichiometric calculation relating organic compound to BaSO₄ precipitate
Calculate mass of organic compound: $\text{Mass} = 2.0 \times 10^{-3} \times 76 = 0.152$ g
How many grams of residue is obtained by heating $2.76$ g of silver carbonate?
(Given : Molar mass of C, O and Ag are $12$, $16$ and $108$ g mol$^{-1}$ respectively)
(1) $1.08$ g
(2) $2.16$ g
(3) $3.24$ g
(4) $4.32$ g
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Answer: Option (2)
Key idea: Thermal decomposition of metal carbonates and stoichiometric calculations
Write the decomposition reaction: $\text{Ag}_2\text{CO}_3 \xrightarrow{\Delta} 2\text{Ag} + \text{CO}_2 + \frac{1}{2}\text{O}_2$. The residue is metallic silver.
Calculate molar mass of $\text{Ag}_2\text{CO}_3$: $M = 2(108) + 12 + 3(16) = 276$ g/mol
Find moles of $\text{Ag}_2\text{CO}_3$: $n = \frac{2.76}{276} = 0.01$ mol
From stoichiometry: $1$ mol $\text{Ag}_2\text{CO}_3$ produces $2$ mol $\text{Ag}$, so $0.01$ mol produces $0.02$ mol $\text{Ag}$
Calculate mass of silver residue: Mass $= 0.02 \times 108 = 2.16$ g
In an estimation of sulphur by Carius method 0.2 g of the substance gave 0.6 g of BaSO$_4$. The percentage of sulphur in the substance is _______%. (Given molar mass in g mol$^{-1}$ S : 32, BaSO$_4$ : 231)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Carius Method for Sulfur Estimation - Stoichiometric calculation relating mass of BaSO₄ precipitate to percentage of sulfur in organic compound
Which of the following contain the same number of atoms?
(Given : Molar mass in g mol$^{-1}$ of H, He, O and S are 1, 4, 16 and 32 respectively)
A. 2 g of O$_2$ gas
B. 4 g of SO$_2$ gas
C. 1400 mL of O$_2$ at STP
D. 0.05 L of He at STP
E. 0.0625 mol of H$_2$ gas
Choose the correct answer from the options given below:
(1) A and B only
(2) B and C only
(3) C and D only
(4) A, C and E only
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Answer: Option (4)
Key idea: Mole concept, Avogadro's number, and calculation of number of atoms from mass, volume at STP, and moles
Match List - I with List - II.
List - I
Mass of substance
A. 1.8 mg water
B. 9.8 mg sulphuric acid
C. 1.8 mg carbon
D. 5.85 mg salt (NaCl)
List - II
Number of atoms
I. $2\times 10^{-4}\times N_A$
II. $1.5\times 10^{-4}\times N_A$
III. $3\times 10^{-4}\times N_A$
IV. $7\times 10^{-4}\times N_A$
Choose the correct answer from the options given below:
(1) A-IV, B-III, C-I, D-II
(2) A-III, B-II, C-IV, D-I
(3) A-III, B-IV, C-II, D-I
(4) A-III, B-IV, C-I, D-II
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Answer: Option (3)
Key idea: Mole concept and calculation of number of atoms from given mass
Calculate moles for each substance using: moles = $\frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
For A (H₂O): moles = $\frac{1.8 \times 10^{-3}}{18} = 10^{-4}$ mol, atoms = $3 \times 10^{-4} \times N_A$ (3 atoms per molecule)
For B (H₂SO₄): moles = $\frac{9.8 \times 10^{-3}}{98} = 10^{-4}$ mol, atoms = $7 \times 10^{-4} \times N_A$ (7 atoms per molecule)
