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NEET • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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61 Hard
61 of 313
Application +4 | -1
If mass percentage of sulphur in a biomolecule 0.02% then the number of sulphur atoms present in 16 g of biomolecule is
View answer and explanation
Correct option: B
6.02×10196.02 \times {10^{19}}
Option B (6.02×10196.02 \times {10^{19}}) is correct because it's calculated by finding the mass of sulfur in 16g of biomolecule and then converting it to the number of sulfur atoms.
62 Easy
62 of 313
Application +4 | -1
The minimum quantity of H2S{\mathrm{H}_2}\mathrm{S} needed to precipitate 63.5 g of Cu2+\mathrm{C}{\mathrm{u}^{2 + }} will be nearly:
View answer and explanation
Correct option: C
34 g
Option C (34 g) is correct because it is the mass of H2SH_2S required to precipitate 63.5 g of Cu2+Cu^{2+} based on stoichiometry.
63 Medium
63 of 313
Remembering +4 | -1
The term atom molecule were introduced by:
View answer and explanation
Correct option: B
Dalton, Avogadro respectively
Option B (Dalton, Avogadro respectively) is correct because Dalton introduced the concept of the atom, and Avogadro contributed significantly to the understanding of molecules.
64 Medium
64 of 313
Analyzing +4 | -1
Assertion: If 200 mL200 \mathrm{~mL} of 0.1 N NaOH0.1 \mathrm{~N} \, \mathrm{NaOH} is added to 200 mL200 \mathrm{~mL} of 0.1 N H2SO40.1 \mathrm{~N} \, \mathrm{H}_2 \mathrm{SO}_4 solution, then the resulting solution is acidic.

Reason: If milliequivalents of acid is greater than milliequivalents of base, then upon mixing the solution is acidic.
View answer and explanation
Correct option: D
Assertion is false, Reason is true
Option D (Assertion is false, Reason is true) is correct. The assertion is false because the milliequivalents of acid and base are equal, resulting in a neutral solution. The reason is true because the acidity of the solution depends on the relative milliequivalents of acid and base.
65 Easy
65 of 313
Analyzing +4 | -1
Equivalent weight of H3PO2\mathrm{H}_3 \mathrm{PO}_2 when it disproportionate into PH3\mathrm{PH}_3 and H3PO3\mathrm{H}_3 \mathrm{PO}_3 is:
View answer and explanation
Correct option: D
3M/43 M / 4
Option D (3M/43 M / 4) is correct because the equivalent weight is calculated by considering both the oxidation and reduction processes in the disproportionation reaction and then averaging the results.
66 Easy
66 of 313
Remembering +4 | -1
How many g are present in one mole of Ag?
View answer and explanation
Correct option: A
107.9
Option A (107.9) is correct because the molar mass of Ag (Silver) is approximately 107.9 g/mol, meaning one mole of Ag weighs 107.9 grams.
67 Easy
67 of 313
Remembering +4 | -1
A molal solution is one that contains one mole of a solute in:
View answer and explanation
Correct option: A
1000 g of the solvent
Option A (1000 g of the solvent) is correct because a molal solution is defined as one mole of solute dissolved in 1000 grams (1 kg) of solvent.
68 Easy
68 of 313
Understanding +4 | -1
The number of units, which are used to express
concentration of solutions from the following
is______

Mass percent, Mole, Mole fraction, Molarity, ppm,
Molality
View answer and explanation
Correct option: A
5
Option A (5) is correct because Mass percent, Mole fraction, Molarity, ppm, and Molality are all units used to express the concentration of solutions.
69 Easy
69 of 313
Understanding +4 | -1
Milliequivalent of a solute in a solution can be given by:
View answer and explanation
Correct option: D
Both (b) and (c )
Option D (Both (b) and (c )) is correct because milliequivalents can be calculated using both N×Vin mLN \times V_{\text{in mL}} and wtEq.wt×1000\frac{{\text{wt}}}{{\text{Eq.wt}}} \times 1000.
70 Medium
70 of 313
Application +4 | -1
What weight of HNO3\mathrm{HN}{\mathrm{O}_3} is needed to convert 5 g I25\,g\,{I_2}  into HIO3,HNO3→NO?\mathrm{HI}{\mathrm{O}_3},\mathrm{HN}{\mathrm{O}_3} \to \mathrm{NO}?
View answer and explanation
Correct option: A
4.13 g
Option A (4.13 g) is correct because it is calculated based on the stoichiometry of the redox reaction and the equivalent weights of the reactants.
71 Hard
71 of 313
Analyzing +4 | -1
Rearrange the following (I to IV) in the order of increasing masses and choose the correct answer from (1), (2), (3) and (4) (Atomic mass: N=14,O=16,Cu=63N=14, O=16, C u=63).

I. 1 molecule of oxygen

II. 1 atom of nitrogen

III. 1×10−10 g1 \times 10^{-10} \mathrm{~g} molecular weight of oxygen

IV. 1×10−10 g1 \times 10^{-10} \mathrm{~g} atomic weight of copper
View answer and explanation
Correct option: A
II < I < III < IV
Option A is correct because it correctly orders the masses from least to greatest: Nitrogen atom < Oxygen molecule < 1e-10 g Oxygen < 1e-10 g Copper.
72 Easy
72 of 313
Analyzing +4 | -1
If the true value for an experimental result is 6.23 and the results reported by three students X, Y and Z are :

X : 6.18 and 6.28

Y : 6.20 and 6.023

Z : 6.22 and 6.24

Which of the following option is correct :
View answer and explanation
Correct option: D
Both X & Y neither precise nor accurate, Z both precise
and accurate.
Option D (Both X & Y neither precise nor accurate, Z both precise and accurate) is correct because X and Y's results are scattered and far from the true value, while Z's results are close to each other and the true value.
73 Hard
73 of 313
Analyzing +4 | -1
The molecular formula of a non-stoichiometric tin oxide containing Sn (II) and Sn (IV) ions is Sn4+4O8\mathrm{Sn}_{4+4} \mathrm{O}_8. Therefore, the molar ratio of Sn (II) to Sn (IV) is approximately
View answer and explanation
Correct option: C
1:41: 4
Option C (1:41: 4) is correct because the formula Sn4.4O8\mathrm{Sn}_{4.4} \mathrm{O}_8 implies that the average oxidation state of Sn is slightly above +2, and the ratio of Sn(II) to Sn(IV) can be calculated based on charge balance.
74 Easy
74 of 313
Application +4 | -1
The formula weight of Al2(SO4)3\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} is 342. A solution containing 342 g of Al2(SO4)3\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} in :
View answer and explanation
Correct option: A
One litre of solution is one molar
Option A (One litre of solution is one molar) is correct because a solution containing the formula weight of a substance in one liter is a 1 M solution.
75 Hard
75 of 313
Analyzing +4 | -1
A mixture of formic acid and oxalic acid is heated with conc. H2SO4.\mathrm{H}_2 \mathrm{SO}_4. The gaseous product is passed into KOH\mathrm{KOH} solution where the volume decreased by 1/6th.1 / 6 \text{th}. What was the molecular proportion of the organic acids, formic and oxalic acid in the mixture?
View answer and explanation
Correct option: B
4:14: 1
Option B (4:1) is correct because the volume decrease in KOH solution corresponds to CO2, and the remaining gas is CO. The ratio of formic acid to oxalic acid is then determined from the amounts of CO and CO2 produced.
76 Medium
76 of 313
Understanding +4 | -1
Which of the following reactions is the metal displacement reaction? Choose the right option.
View answer and explanation
Correct option: D
Cr2O3+2Al→ΔAl2O3+2CrC{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr
Option D (Cr2O3+2Al→ΔAl2O3+2CrC{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr) is correct because aluminum displaces chromium from its oxide, which is a metal displacement reaction.
77 Hard
77 of 313
Remembering +4 | -1
In the given chemical reaction, colours of the Fe2+  and  Fe3+\mathrm{F}{\mathrm{e}^{\mathrm{2 + }}}\,\,\mathrm{and}\,\,\mathrm{ F}{\mathrm{e}^{\mathrm{3 + }}} ions, are respectively
5Fe2++MnO4−+8H+→Mn2++4H2O+5Fe3+\mathrm{5}F{e^{\mathrm{2 + }}}\mathrm{ + }M\mathrm{n}O_4^ - + 8{H^ + } \to M{n^{2 + }} + 4{H_2}O + 5F{e^{3 + }}
View answer and explanation
Correct option: D
green, yellow
Option D is correct because Fe2+Fe^{2+} ions are typically green, and Fe3+Fe^{3+} ions are typically yellow in solution.
78 Easy
78 of 313
Analyzing +4 | -1
Which sample contains the smallest number of atoms?
View answer and explanation
Correct option: D
2.5 mol Au2.5 \,\mathrm{mol} \,\mathrm{Au}
Option D (2.5 mol Au2.5 \,\mathrm{mol} \,\mathrm{Au}) is correct because it contains the smallest number of atoms, calculated as 2.5 mol * 1 atom/mol = 2.5 * Avogadro's number of atoms.
79 Easy
79 of 313
Understanding +4 | -1
The formation of COCO and CO2C{O_2} illustrates the law of
View answer and explanation
Correct option: C
Multiple proportion
The correct answer is C (Multiple proportion) because carbon and oxygen form two different compounds, CO and CO2, where the ratio of oxygen masses combining with a fixed mass of carbon is a simple whole number.
80 Medium
80 of 313
Application +4 | -1
Weight of 112  mL112\,\,mL of oxygen at NTP on liquification would be:
View answer and explanation
Correct option: D
0.16  g0.16\,\,g
Option D (0.16 g0.16\,g) is correct because the weight of oxygen is calculated using the molar mass and the volume at NTP.
81 Medium
81 of 313
Analyzing +4 | -1
Which of the following statement is correct
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the statements are true: 28g CO contains 12g C and 16g O, one mole of CO reacts completely with half mole of O2 to form CO2, and N2 and CO have the same molar mass (28 g/mol).
82 Easy
82 of 313
Analyzing +4 | -1
Assertion: Density == specific gravity ×0.99823\times 0.99823 at 20∘C20^{\circ} \mathrm{C}.
Reason: Density == specific gravity at 4∘C4^{\circ} \mathrm{C}.
View answer and explanation
Correct option: D
Both Assertion and Reason are correct but Reason is not correct explanation of Assertion
Option D (Both Assertion and Reason are correct but Reason is not correct explanation of Assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
83 Easy
83 of 313
Application +4 | -1
10  g  CaCO310\;g\;\mathrm{CaC}{\mathrm{O}_3} on heating gives 5.6 g CaO and &hellip;. g CO2.\mathrm{C}{\mathrm{O}_2}.
View answer and explanation
Correct option: A
4.4
Option A (4.4) is correct because the mass of CO2CO_2 produced from the decomposition of CaCO3CaCO_3 is 4.4g, following stoichiometry.
84 Easy
84 of 313
Analyzing +4 | -1
Assertion: Equivalent weight of NH3\mathrm{NH}_3 in the reaction N2→NH3\mathrm{N}_2 \rightarrow \mathrm{NH}_3 is 17/317 / 3 while that of N2\mathrm{N}_2 is 28/6.28/6.

Reason: Equivalent weight = Molecular weightnumber of e− lost or gained \frac{\text {Molecular weight}}{\text {number of } e^- \text{ lost or gained }}.
View answer and explanation
Correct option: A
Assertion is true, Reason is true and Reason is correct explanation for Assertion
Option A (Assertion is true, Reason is true and Reason is correct explanation for Assertion) is correct because the equivalent weight is indeed the molecular weight divided by the number of electrons lost or gained, and both the assertion and reason are correct and related.
85 Medium
85 of 313
Understanding +4 | -1
Which of the following is correct for
C(graphite)+O2(gas)→CO2,heat=348  kJ?\mathrm{C}\left( {\mathrm{graphite}} \right) + {\mathrm{O}_2}\left( {\mathrm{gas}} \right) \to \mathrm{C}{\mathrm{O}_2},heat\mathrm{ } = 348\;kJ?
View answer and explanation
Correct option: D
Mass of product = Mass of reactant
The correct answer is D (Mass of product = Mass of reactant) because according to the law of conservation of mass, mass is neither created nor destroyed in a chemical reaction.
86 Hard
86 of 313
Analyzing +4 | -1
Assertion : The activity of 1g1 g pure uranium-235 will be greater than the same amount present as U3O8U_3 O_8.

Reason : In the combined state, the activity of the radioactive element decreases.
View answer and explanation
Correct option: D
If the assertion and reason both are false.
Option D is correct because both the assertion and the reason are false. The activity of a radioactive element depends on its specific activity and amount, not its chemical state.
87 Hard
87 of 313
Application +4 | -1
A gas has a vapour density 11.2. The volume occupied by 1g of the gas at NTP is
View answer and explanation
Correct option: A
1 L
Option A (1 L) is correct because the volume occupied by 1g of the gas at NTP is calculated using the vapor density and molar volume.
88 Easy
88 of 313
Application +4 | -1
Volume occupied by 3.01×1023    3.01 \times {10^{23}}\;\; molecules of acetylene at NTP is:
View answer and explanation
Correct option: B
11.2  litre11.2\,\,\mathrm{litre}
Option B (11.2 litre) is correct because it is half the molar volume at NTP, corresponding to half a mole of acetylene.
89 Hard
89 of 313
Analyzing +4 | -1
Which of the given reactions is not an example of disproportionation reaction?
View answer and explanation
Correct option: C
MnO4−+4H++3e−→MnO2+2H2O\mathrm{MnO}_4^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}
Option C is correct because disproportionation involves a single element undergoing both oxidation and reduction, which is not happening in option C where electrons are simply added to MnO4-.
90 Easy
90 of 313
Remembering +4 | -1
Which property of an element is not variable?
View answer and explanation
Correct option: B
At. wt.
Option B (At. wt.) is correct because the atomic weight of an element is a constant property, while valence and equivalent weight can vary depending on the chemical context.
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