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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Hard

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31 Hard
31 of 77
Analyzing +4 | -1
One mole of mixture of N2,NO2\mathrm{N}_2, \mathrm{NO}_2 and N2O4\mathrm{N}_2 \mathrm{O}_4 has a mean molar mass of 55.4 g.55.4 \mathrm{~g}. On heating to a temperature, at which all the N2O4\mathrm{N}_2 \mathrm{O}_4 is dissociated into NO2,\mathrm{NO}_2, the mean molar mass tends to a lower value of 39.6 g.39.6 \mathrm{~g}. What is the mole ratio of N2,NO2\mathrm{N}_2, \mathrm{NO}_2 and N2O4\mathrm{N}_2 \mathrm{O}_4 in the original mixture?
View answer and explanation
Correct option: A
5:1:45: 1: 4
Option A (5: 1: 4) is correct because solving the system of equations derived from the given molar mass data yields the mole ratio of N2, NO2, and N2O4 as 5:1:4.
32 Hard
32 of 77
Application +4 | -1
What will be the minimum molar mass of a compound containing 0.39%0.39 \% of nickel?
View answer and explanation
Correct option: B
15 kg
Option B (15 kg) is correct because it is calculated by assuming the compound contains at least one nickel atom and then determining the molar mass based on the given percentage.
33 Hard
33 of 77
Application +4 | -1
The number of moles of KMnO4KMn{O_4} that will be needed to react completely with one mole of oxalate ion in acidic solution is
View answer and explanation
Correct option: C
25\frac{2}{5}
The correct answer is C (2/5) because the stoichiometry of the redox reaction between KMnO4KMn{O_4} and oxalate ion in acidic solution requires 2 moles of KMnO4KMn{O_4} for every 5 moles of oxalate.
34 Hard
34 of 77
Application +4 | -1
An unknown compound A\mathrm{A} dissociates at 500∘C500^{\circ} \mathrm{C} to give products as follows A(g)⇄B(g)+C(g)+D(g)A(g) \rightleftarrows B(g)+C(g)+D(g) Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10%10 \%. What will be the molecular weight of compound A-
View answer and explanation
Correct option: A
120
Option A (120) is correct because the molecular weight of compound A is calculated using the vapour density at equilibrium and the degree of dissociation.
35 Hard
35 of 77
Analyzing +4 | -1
A mixture of NaHC2O4\mathrm{NaHC}_2 \mathrm{O}_4 and KHC2O4⋅H2C2O4\mathrm{KHC}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 required equal volumes of 0.2 N KMnO40.2 \mathrm{~N} \mathrm{~KMnO}_4 and 0.12 NNaOH0.12 \mathrm{~N} \mathrm{NaOH} separately. What is the molar ratio of NaHC2O4\mathrm{NaHC}_2 \mathrm{O}_4 and KHC2O4⋅H2C2O4\mathrm{KHC}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 in the mixture?
View answer and explanation
Correct option: D
3:13: 1
Option D (3:1) is correct because by setting up equations based on the stoichiometry of the reactions with KMnO4 and NaOH, we can solve for the molar ratio of NaHC2O4 and KHC2O4·H2C2O4.
36 Hard
36 of 77
Application +4 | -1
The sulphate of an element contains 42.2%42.2 \% element. The equivalent weight of the metal would be
View answer and explanation
Correct option: B
35
Option B (35) is correct because the equivalent weight is calculated by finding the mass of the element and dividing by its valency, which is 2 in this case.
37 Hard
37 of 77
Application +4 | -1
Two elements XX and YY have atomic weights of 14 and 16. They form a serics of compounds A,B,C,DA, B, C, D and EE in which the same amount of element X,Y{X},{Y} is present in the ratio 1:2:3:4:5.1: 2: 3: 4: 5. If the compound AA has 28 parts by weight of XX and 16 parts by weight of Y,{Y}, then the compound of C{C} will have 28 parts weight of X{X} and
View answer and explanation
Correct option: B
48 parts by weight of Y{Y}
Option B (48 parts by weight of Y) is correct because if compound A is XY, then compound C (XY3) will have 28 parts by weight of X and 48 parts by weight of Y.
38 Hard
38 of 77
Application +4 | -1
0.40 g0.40 \mathrm{~g} of helium in a bulb at a temperature of T KT \mathrm{~K} had a pressure of pp atm. When the bulb was immersed in water bath at temperature 50 K50 \mathrm{~K} more than the first one, 0.08 g0.08 \mathrm{~g} of gas had to be removed to restore the original pressure. TT is
View answer and explanation
Correct option: D
200 K200 \mathrm{~K}
Option D (200 K) is correct because using the ideal gas law relationship, the temperature T is found to be 200 K.
39 Hard
39 of 77
Application +4 | -1
Calculate the molality of 1 litre solution of 93% H2SO493 \% \, \mathrm{H}_2 \mathrm{SO}_4 (weight/volume). The density of the solution is 1.84 g/mL1.84 \mathrm{~g} / \mathrm{mL}.
View answer and explanation
Correct option: C
10.42 m
Option C (10.42 m) is correct because the molality is calculated using the mass of solute and solvent in the solution.
40 Hard
40 of 77
Analyzing +4 | -1
Iron forms two oxides. If for the same mass of iron, the mass of oxygen combined in the first oxide is two-third of the mass of oxygen combined in the second oxide, then the ratio of valency of iron in first and second oxide is
View answer and explanation
Correct option: B
2:32: 3
Option B (2:32: 3) is correct because the ratio of oxygen masses is 2:3, implying the ratio of iron valencies is also 2:3 based on the oxide formulas.
41 Hard
41 of 77
Application +4 | -1
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2{O_2} by volume for complete combustion. After combustion, the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is
View answer and explanation
Correct option: A
C2H12 {C_2}{H_{12}}\,
The correct answer is A (C2H12 {C_2}{H_{12}}\,) because the hydrocarbon formula is determined by using stoichiometry of the combustion reaction and given volume data.
42 Hard
42 of 77
Analyzing +4 | -1
For the reaction,

FeS2+KMnO4+H∗⟶Fe3++H2SO4+Mn2++H2O\mathrm{FeS}_2+\mathrm{KMnO}_4+\mathrm{H}^* \longrightarrow \mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}

if the molar mass of FeS2\mathrm{FeS}_2 is M\mathrm{M}, then equivalent mass of FeS2\mathrm{FeS}_2 would be equal to
View answer and explanation
Correct option: D
M15\frac{M}{15}
Option D (M15\frac{M}{15}) is correct because the n-factor for FeS2 in the given redox reaction is 15, so the equivalent mass is M/15.
43 Hard
43 of 77
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 (II < I < III < IV) is correct because it represents the increasing order of masses calculated for each species.
44 Hard
44 of 77
Application +4 | -1
How many litres of detonating gas may be produced at 0∘C0^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} from the decomposition of 0.1 mole of water by an electric current?
View answer and explanation
Correct option: C
3.36 L3.36 \mathrm{~L}
Option C (3.36 L) is correct because the volume of detonating gas (H₂ and O₂) produced from the decomposition of 0.1 mole of water is calculated using stoichiometry and the ideal gas law.
45 Hard
45 of 77
Application +4 | -1
25 ml25 \mathrm{~ml} solution containing (6.35 g/L\left(6.35 \mathrm{~g} / \mathrm{L}\right. of KH3(C2O4)2⋅2H2O\mathrm{K} \mathrm{H}_3\left(\mathrm{C}_2 \mathrm{O}_4\right)_2 \cdot 2 \mathrm{H}_2 \mathrm{O} required V1ml\mathrm{V}_1 \mathrm{ml} of 0.1 N0.1 \mathrm{~N} NaOH\mathrm{NaOH} and V2ml\mathrm{V}_2 \mathrm{ml} of 0.1 N KMnO40.1 \mathrm{~N} \mathrm{~KMnO}_4 in two separate titrations. Calculate V1\mathrm{V}_1 and V2\mathrm{V}_2.
KH3(C2O4)2⋅2H2O;M0=254 gm/mol\mathrm{KH}_3\left(\mathrm{C}_2 \mathrm{O}_4\right)_2 \cdot 2 \mathrm{H}_2 \mathrm{O} ; \mathrm{M}_0=254 \mathrm{~gm} / \mathrm{mol}
View answer and explanation
Correct option: A
18.75 ml18.75 \mathrm{~ml} , 25 ml25 \mathrm{~ml}
Option A (18.75 ml , 25 ml) is correct because it calculates the volumes of NaOH and KMnO4 required based on the given normality and the amount of the compound.
46 Hard
46 of 77
Application +4 | -1
One g\mathrm{g} of impure Na2CO3\mathrm{Na}_2 \mathrm{CO}_3 is dissolved in water and the solution is made upto 250 mL250 \mathrm{~mL}. To 50 mL50 \mathrm{~mL} of this made up solution, 50 mL50 \mathrm{~mL} of 0.1 N HCl0.1 \mathrm{~N} \mathrm{~HCl} is added and the mixture after shaking well required 10 mL10 \mathrm{~mL} of 0.16 N0.16 \mathrm{~N} sodium hydroxide solution for complete neutralization. Calculate the per cent purity of the sample of Na2CO3\mathrm{Na}_2 \mathrm{CO}_3.
View answer and explanation
Correct option: A
90%90 \%
Option A (90%90 \%) is correct because the purity is calculated based on the amount of Na2CO3Na_2CO_3 that reacts with HCl and NaOH.
47 Hard
47 of 77
Application +4 | -1
A 0.518 g sample of lime stone is dissolved in HClHCl and then the calcium is precipitated as CaC2O4\mathrm{Ca}{\mathrm{C}_2}{\mathrm{O}_4}. After filtering and washing the precipitate, it requires 40.0 mL of 0.250NKMnO4,0.250N\mathrm{KMn}{\mathrm{O}_4},, solution acidified with H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} to titrate is as, MnO4−+H++C2O42−→Mn2++CO2+2H2O\mathrm{MnO}_4^ - + {\mathrm{H}^ + } + {\mathrm{C}_2}\mathrm{O}_4^{2 - } \to \mathrm{M}{\mathrm{n}^{2 + }} + \mathrm{C}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}. The percentage of CaO\mathrm{CaO} in the sample is :
View answer and explanation
Correct option: A
54.0%54.0\mathrm{ }\%
Option A (54.0%54.0\%) is correct because the percentage of CaO is calculated by equating the meq of limestone, CaC2O4, KMnO4 and CaO.
48 Hard
48 of 77
Application +4 | -1
Calculate the molality of 1 litre solution of 93% H2SO493 \% \, \mathrm{H}_2 \mathrm{SO}_4 (weight/volume). The density of the solution is 1.84 g/mL1.84 \mathrm{~g} / \mathrm{mL}.
View answer and explanation
Correct option: C
10.42 m
Option C (10.42 m) is correct because the molality is calculated based on the mass of solute and solvent in the solution.
49 Hard
49 of 77
Application +4 | -1
1 mol each of H3PO2,H3PO3{H}_3 {PO}_2, {H}_3 {PO}_3 and H3PO4{H}_3 {PO}_4 will neutralize x molx \mathrm{~mol} of NaOH,y mol{NaOH}, y \mathrm{~mol} of Ca(OH)2{Ca}({OH})_2 and z mol{z} \mathrm{~mol} of Al(OH)3.x,y{Al}({OH})_3 . x, y and zz are in the ratio
View answer and explanation
Correct option: D
1:1:11: 1: 1
Option B (4.78×10−11 mol L−14.78 \times 10^{-11} \mathrm{~mol} \mathrm{~L}^{-1}) is correct because it is the concentration of CO3-2 calculated using the given pH, K1, and K2 values for carbonic acid.
50 Hard
50 of 77
Application +4 | -1
The density of 3M3 \mathrm{M} solution of sodium chloride is 1.252 g mL−11.252 \mathrm{~g} \mathrm{~mL}^{-1}. The molality of the solution will be:

(molar mass, NaCl=58.5 g mol−1\mathrm{NaCl}=58.5 \mathrm{~g} \mathrm{~mol}^{-1} )
View answer and explanation
Correct option: C
2.79 m2.79 \mathrm{~m}
Option C (2.79 m) is correct because it is calculated using the relationship between molarity, molality, and density.
51 Hard
51 of 77
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 (green, yellow) is correct because Fe2+\mathrm{F}{\mathrm{e}^{\mathrm{2 + }}} ions are typically green in solution, while Fe3+\mathrm{F}{\mathrm{e}^{\mathrm{3 + }}} ions are typically yellow.
52 Hard
52 of 77
Application +4 | -1
Chlorofluorocarbons such as CCl3F(M=137.5){CCl}_3 {F}(\mathrm{M}=137.5) and CCl2F2(M=121){CCl}_2 {F}_2(\mathrm{M}=121) have been linked to ozone depletion in Antarctica. As of 2004, these gases were found in 275 and 605 parts per trillion (1012),\left(10^{12}\right), by volume. What are the concentrations of these gases under conditions typical of Antarctica stratosphere (200 K(200 \mathrm{~K} and 0.08 atm)?(R=0.08 l-atm/K-mol)0.08 \mathrm{~atm}) ?(R=0.08 \, l \text{-atm/K-mol})
View answer and explanation
Correct option: A
[CCl3F]=1.375×10−12 mol l−1,[CCl2F2]=3.025×10−12 mol l−1\left[{CCl}_3 {F}\right]=1.375 \times 10^{-12} \mathrm{~mol} \, {l}^{-1},\left[{CCl}_2 {F}_2\right]=3.025 \times 10^{-12} \mathrm{~mol} \, l^{-1}
Option A ([CCl3F]=1.375×10−12 mol,l−1,[CCl2F2]=3.025×10−12 mol,l−1\left[{CCl}_3 {F}\right]=1.375 \times 10^{-12} \mathrm{~mol} \\, {l}^{-1},\left[{CCl}_2 {F}_2\right]=3.025 \times 10^{-12} \mathrm{~mol} \\, l^{-1}) is correct because it correctly calculates the concentrations using the ideal gas law and the given parts per trillion values.
53 Hard
53 of 77
Application +4 | -1
The number of gram atoms of oxygen present in 0.3 gram mole of (COOH)2⋅2H2O({COOH})_2 \cdot 2 {H}_2 {O} is
View answer and explanation
Correct option: B
1.8
Option B (1.8) is correct because it's the number of gram atoms of oxygen calculated from the given moles of the compound.
54 Hard
54 of 77
Application +4 | -1
The mineral haematite is Fe2O3\mathrm{Fe}_2 \mathrm{O}_3. Haematite ore contains unwanted material called gangue in addition of Fe2O3\mathrm{Fe}_2 \mathrm{O}_3. If 5.0 kg5.0 \mathrm{~kg} of ore contains 2.78 kg2.78 \mathrm{~kg} of Fe\mathrm{Fe}, what per cent of ore is Fe2O3\mathrm{Fe}_2 \mathrm{O}_3?
View answer and explanation
Correct option: A
79.4%79.4 \%
Option A (79.4 %) is correct because it accurately calculates the percentage of Fe2O3 in the ore based on the given mass of Fe.
55 Hard
55 of 77
Application +4 | -1
If the weight of metal oxide is xg{xg} containing yg{yg} of oxygen, the equivalent weight of metal will be
View answer and explanation
Correct option: D
E=8(x−y)yE=\frac{8(x-y)}{y}
Option D (E=8(x−y)yE=\frac{{8(x-y)}}{{y}}) is correct because it represents the equivalent weight of the metal calculated from the ratio of the weight of the metal to the weight of oxygen multiplied by 8.
56 Hard
56 of 77
Application +4 | -1
If mAm_A gram of a metal AA displaces mBm_B gram of another metal BB from its salt solution and if the equivalent weights are EAE_A and EBE_B respectively then equivalent weight of AA can be expressed as:
View answer and explanation
Correct option: A
EA=mAmB×EBE_A=\frac{m_A}{m_B} \times E_B
Option A (EA=mAmB×EBE_A=\frac{m_A}{m_B} \times E_B) is correct because it directly applies Faraday's laws of electrolysis, which state that the mass of a substance deposited or dissolved is proportional to its equivalent weight.
57 Hard
57 of 77
Application +4 | -1
40ml40 \mathrm{ml} of 0.8 N0.8 \mathrm{~N} oxalic acid dihydrate is shaken with 0.75 g0.75 \mathrm{~g} active charcoal. The final concentration of the solution after adsorption is 0.2 N0.2 \mathrm{~N}. What is the amount of oxalic acid adsorbed per gm of carbon?
View answer and explanation
Correct option: D
2.02 g2.02 \mathrm{~g}
Option D (2.02 g2.02 \mathrm{~g}) is correct because it is the amount of oxalic acid adsorbed per gram of carbon, calculated from the change in concentration.
58 Hard
58 of 77
Application +4 | -1
What is the mass of 3.01×10223.01 \times 10^{22} molecules of ammonia?
View answer and explanation
Correct option: A
0.85 g
Option A (0.85 g) is correct because it is calculated using the mole concept and Avogadro's number to find the mass of the given number of ammonia molecules.
59 Hard
59 of 77
Analyzing +4 | -1
Cupric ammonium sulphate was found to contain
27.03%27.03\% water of crystallization by mass. Upon
strongly heating, it gave cupric oxide corresponding
to 19.89%19.89\% of starting mass. Find the empirical
formula of cupric ammonium sulphate. (Cu=63.5)(\text{Cu} = 63.5)
View answer and explanation
Correct option: A
CuSO4⋅(NH4)2SO4⋅6H2O\mathrm{CuSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}
Option A is correct because the empirical formula is determined by calculating the moles of each component and finding the simplest whole number ratio, which corresponds to CuSO₄⋅(NH₄)₂SO₄⋅6H₂O.
60 Hard
60 of 77
Application +4 | -1
Mole fraction of solute in an aqueous solution of NaOH\mathrm{NaOH} is 0.1.0.1. If the specific gravity of the solution is 1.4,1.4, then the molarity of the solution is
View answer and explanation
Correct option: A
6.93
Option A (6.93) is correct because the molarity is calculated using the mole fraction, specific gravity, and molar masses of the solute and solvent.
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