Some Basic Concepts of Chemistry
247 Questions
Start JEE Mains Test
2021
Q151
JEE Mains
Numerical
14 Mar 2026
The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4.2H2O) in 250 mL of water in mol L$-$1 is x $\times$ 10$-$2. The value of x is _____________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]
Correct Answer: 20
Explanation:
$[{H_2}{C_2}{O_4}.2{H_2}O] = {{weight/{M_w}} \over {V(L)}}$
$ \Rightarrow x \times {10^{ - 2}} = {{6.3/126} \over {250/1000}}$
$x = 20$
$ \Rightarrow x \times {10^{ - 2}} = {{6.3/126} \over {250/1000}}$
$x = 20$
2021
Q152
JEE Mains
Numerical
14 Mar 2026
100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x $\times$ 10$-$2. The value of x is ____________. (Nearest integer)
[Atomic weight : H = 1.008; C = 12.00; O = 16.00]
[Atomic weight : H = 1.008; C = 12.00; O = 16.00]
Correct Answer: 19
Explanation:
C3H8(g) + 5O2(g) $\to$ 3CO2(g) + 4H2O(l)
mole fraction of CO2 in the final reaction mixture (heterogenous)
${X_{C{O_2}}} = {{6.81} \over {19.9 + 6.81 + 9.08}}$
= 0.1902 = 19.02 $\times$ 10$-$2
$\Rightarrow$ 19
| t = 0 | 2.27 mole | 31.25 mol | ||
|---|---|---|---|---|
| t = $\infty $ | 0 | 19.9 mol | 6.81 mol | 9.08 mol |
mole fraction of CO2 in the final reaction mixture (heterogenous)
${X_{C{O_2}}} = {{6.81} \over {19.9 + 6.81 + 9.08}}$
= 0.1902 = 19.02 $\times$ 10$-$2
$\Rightarrow$ 19
2021
Q153
JEE Mains
Numerical
14 Mar 2026
In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ______________. (Nearest integer)
[Atomic mass : Ag = 108, Br = 80]
[Atomic mass : Ag = 108, Br = 80]
Correct Answer: 40
Explanation:
nAgBr = ${{0.188g} \over {188g/mol}}$ = 10$-$3 mol
$\Rightarrow$ nBr = nAgBr = 0.001 mol
$\Rightarrow$ massBr = (0.001 $\times$ 80) gm = 0.08 gm
$\Rightarrow$ mass% = ${{0.08 \times 100} \over {0.2}} = 40\% $
$\Rightarrow$ nBr = nAgBr = 0.001 mol
$\Rightarrow$ massBr = (0.001 $\times$ 80) gm = 0.08 gm
$\Rightarrow$ mass% = ${{0.08 \times 100} \over {0.2}} = 40\% $
2021
Q154
JEE Mains
Numerical
14 Mar 2026
100 mL of Na3PO4 solution contains 3.45 g of sodium. The molarity of the solution is _____________ $\times$ 10$-$2 mol L$-$1. (Nearest integer)
[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
Correct Answer: 50
Explanation:
Molarity of Na3PO4 Solution = ${{{n_{N{a_3}P{O_4}}}} \over {volume\,of\,solution\,in\,L}}$
$ = {{{1 \over 3} \times {{3.45} \over {23}}mol} \over {0.1\,L}}$
= 0.5 = 50 $\times$ 10$-$2
$ = {{{1 \over 3} \times {{3.45} \over {23}}mol} \over {0.1\,L}}$
= 0.5 = 50 $\times$ 10$-$2
2021
Q155
JEE Mains
Numerical
14 Mar 2026
An aqueous KCl solution of density 1.20 g mL$-$1 has a molality of 3.30 mol kg$-$1. The molarity of the solution in mol L$-$1 is ____________ (Nearest integer) [Molar mass of KCl = 74.5]
Correct Answer: 3
Explanation:
1000 kg solvent has 3.3 moles of KCl
1000 kg solvent $\to$ 3.3 $\times$ 74.5 gm KCl $\to$ 245.85
Weight of solution = 1245.85 gm
Volume of solution = ${{1245.85} \over {1.2}}$ ml
So molarity = ${{3.3 \times 1.2} \over {1245.85}} \times 1000 = 3.17$
1000 kg solvent $\to$ 3.3 $\times$ 74.5 gm KCl $\to$ 245.85
Weight of solution = 1245.85 gm
Volume of solution = ${{1245.85} \over {1.2}}$ ml
So molarity = ${{3.3 \times 1.2} \over {1245.85}} \times 1000 = 3.17$
2021
Q156
JEE Mains
Numerical
14 Mar 2026
2SO2(g) + O2(g) $\to$ 2SO3(g)
The above reaction is carried out in a vessel starting with partial pressure PSO2 = 250 m bar, PO2 = 750 m bar and PSO3 = 0 bar. When the reaction is complete, the total pressure in the reaction vessel is _______ m bar. (Round off of the nearest integer).
The above reaction is carried out in a vessel starting with partial pressure PSO2 = 250 m bar, PO2 = 750 m bar and PSO3 = 0 bar. When the reaction is complete, the total pressure in the reaction vessel is _______ m bar. (Round off of the nearest integer).
Correct Answer: 875
Explanation:
2SO2(g) + O2(g) $\to$ 2SO3(g)
$\therefore$ Final total pressure = 625 + 250 = 875 m bar
$\therefore$ Final total pressure = 625 + 250 = 875 m bar
2021
Q157
JEE Mains
Numerical
14 Mar 2026
The density of NaOH solution is 1.2 g cm$-$3. The molality of this solution is _____________ m. (Round off to the Nearest Integer)
[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H2O : 1.0 g cm$-$3]
[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H2O : 1.0 g cm$-$3]
Correct Answer: 5
Explanation:
Consider 1 litre solution
mass of solution = (1.2 $\times$ 1000)g = 1200 gm
Neglecting volume of NaOH
Mass of water = 1000 gm
$\Rightarrow$ Mass of NaOH = (1200 $-$ 1000)gm = 200 gm
$\Rightarrow$ Moles of NaOH = ${{200g} \over {50g/mol}} = 5$ mol
$\Rightarrow$ molality = ${{5mol} \over {1kg}} = 5$ m
mass of solution = (1.2 $\times$ 1000)g = 1200 gm
Neglecting volume of NaOH
Mass of water = 1000 gm
$\Rightarrow$ Mass of NaOH = (1200 $-$ 1000)gm = 200 gm
$\Rightarrow$ Moles of NaOH = ${{200g} \over {50g/mol}} = 5$ mol
$\Rightarrow$ molality = ${{5mol} \over {1kg}} = 5$ m
2021
Q158
JEE Mains
Numerical
14 Mar 2026
The number of significant figures in 0.00340 is ____________.
Correct Answer: 3
Explanation:
Number of significant figures = 3
2021
Q159
JEE Mains
Numerical
14 Mar 2026
When 10 mL of an aqueous solution of Fe2+ ions was titrated in the presence of dil H2SO4 using diphenylamine indicator, 15 mL of 0.02 M solution of K2Cr2O7 was required to get the end point. The molarity of the solution containing Fe2+ ions is x $\times$ 10$-$2 M. The value of x is __________. (Nearest integer)
Correct Answer: 18
Explanation:
milli-equivalents of Fe2+ = milli-equivalents of K2Cr2O7
M $\times$ 10 $\times$ 1 = 0.02 $\times$ 15 $\times$ 6
$ \Rightarrow $ M = 0.18 = 18 $\times$ 10$-$2 M
M $\times$ 10 $\times$ 1 = 0.02 $\times$ 15 $\times$ 6
$ \Rightarrow $ M = 0.18 = 18 $\times$ 10$-$2 M
2021
Q160
JEE Mains
Numerical
14 Mar 2026
Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is ___________ $\times$ 10$-$1 g. (in nearest integer)
Correct Answer: 464
Explanation:
${C_4}{H_{10}} + {{13} \over 2}{O_2}\buildrel {} \over
\longrightarrow 4C{O_2} + 5{H_2}O$
Moles of ${H_2}O = {{72} \over {18}} = 4$
Moles of C4H10 used $ = {1 \over 5} \times 4$
Wight of C4H10 used $ = {4 \over 5} \times 58$
= 46.4 gm = 464 $ \times $ 10-1 g
Moles of ${H_2}O = {{72} \over {18}} = 4$
Moles of C4H10 used $ = {1 \over 5} \times 4$
Wight of C4H10 used $ = {4 \over 5} \times 58$
= 46.4 gm = 464 $ \times $ 10-1 g
2021
Q161
JEE Mains
Numerical
14 Mar 2026
If the concentration of glucose (C6H12O6) in blood is 0.72 g L$-$1, the molarity of glucose in blood is ____________ $\times$ 10$-$3 M. (Nearest integer)
[Given : Atomic mass of C = 12, H = 1, O = 16 u]
[Given : Atomic mass of C = 12, H = 1, O = 16 u]
Correct Answer: 4
Explanation:
[Glucose] = ${{C(gm/l)} \over {M(gm/mol)}} = {{0.72} \over {180}} = 4 \times {10^{ - 3}}$ M
2021
Q162
JEE Mains
Numerical
14 Mar 2026
4g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g of Na2CO3. The value of x is ____________ g. (Nearest integer)
Correct Answer: 1
Explanation:
Mass of NaOH = x
Moles of NaOH = ${x \over {40}}$
Mass of Na2CO3 = y
Moles of Na2CO3 = ${y \over {106}}$
${x \over {40}} = {y \over {106}}$
x + y = 4
x = 1.1, y = 2.9
x = 1.1 $ \approx $ 1 (nearest integer)
Moles of NaOH = ${x \over {40}}$
Mass of Na2CO3 = y
Moles of Na2CO3 = ${y \over {106}}$
${x \over {40}} = {y \over {106}}$
x + y = 4
x = 1.1, y = 2.9
x = 1.1 $ \approx $ 1 (nearest integer)
2021
Q163
JEE Mains
Numerical
14 Mar 2026
250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ___________ $\times$ 1021. (Nearest integer)
(NA = 6.022 $\times$ 1023)
(NA = 6.022 $\times$ 1023)
Correct Answer: 226
Explanation:
We know that, number of moles = VL $\times$ molarity and number of millimoles = VmL $\times$ molarity
So, millimoles of NaOH = 250 $\times$ 0.5 = 125
Millimoles of HCl = 500 $\times$ 1 = 500
Now, reaction is
125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375
Moles of HCl = 375 $\times$ 10$-$3
Number of HCl molecules
= Avogadro's constant (NA) $\times$ moles of HCl
= 6.022 $\times$ 1023 $\times$ 375 $\times$ 10$-$3
= 225.8 $\times$ 1021 = 226 $\times$ 1021
Therefore, answer is 226.
So, millimoles of NaOH = 250 $\times$ 0.5 = 125
Millimoles of HCl = 500 $\times$ 1 = 500
Now, reaction is
125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375
Moles of HCl = 375 $\times$ 10$-$3
Number of HCl molecules
= Avogadro's constant (NA) $\times$ moles of HCl
= 6.022 $\times$ 1023 $\times$ 375 $\times$ 10$-$3
= 225.8 $\times$ 1021 = 226 $\times$ 1021
Therefore, answer is 226.
2021
Q164
JEE Mains
Numerical
14 Mar 2026
An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass = 180.0 g mol$-$1) needed to meet this energy requirement is ____________ g.
(Use : $\Delta$CH(glucose) = $-$2700 kJ mol$-$1)
(Use : $\Delta$CH(glucose) = $-$2700 kJ mol$-$1)
Correct Answer: 667
Explanation:
1 mole glucose give 2700 kJ energy,
so mole of glucose needed for 104 kJ energy
= ${{10000} \over {2700}} = 3.703$ moles
Weight of glucose = 3.703 $\times$ 180 g/moles
= 666.666
$\approx$ 667 g
Hence, amount of glucose required is 667 g.
so mole of glucose needed for 104 kJ energy
= ${{10000} \over {2700}} = 3.703$ moles
Weight of glucose = 3.703 $\times$ 180 g/moles
= 666.666
$\approx$ 667 g
Hence, amount of glucose required is 667 g.
2021
Q165
JEE Mains
Numerical
14 Mar 2026
10.0 mL of Na2CO3 solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings :
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___________ mM.
(Round off to the Nearest Integer).
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___________ mM.
(Round off to the Nearest Integer).
Correct Answer: 50
Explanation:
From the given value of HCl, it is clear that most appropriate volume of HCl used is 5 ml because it occurs most number of times.
Na2CO3 + 2 HCl $ \to $ 2 NaCl + CO2 + H2O
n factor of Na2CO3 = 2
n factor of HCl = 1
equivalent of Na2CO3 = equivalent of HCl
$ \Rightarrow $ ${{10} \over {1000}} \times 2 \times M = {5 \over {1000}} \times 1 \times 0.2$
$ \Rightarrow M = {1 \over {20}}M$
= 0.05 M
= 50 $\times$ 10$-$3 M
= 50 mM
Na2CO3 + 2 HCl $ \to $ 2 NaCl + CO2 + H2O
n factor of Na2CO3 = 2
n factor of HCl = 1
equivalent of Na2CO3 = equivalent of HCl
$ \Rightarrow $ ${{10} \over {1000}} \times 2 \times M = {5 \over {1000}} \times 1 \times 0.2$
$ \Rightarrow M = {1 \over {20}}M$
= 0.05 M
= 50 $\times$ 10$-$3 M
= 50 mM
2021
Q166
JEE Mains
Numerical
14 Mar 2026
__________ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) (C3H4O) if the percentage yield is 64. (Round off to the Nearest Integer).
[Given : Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u ]
[Given : Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u ]
Correct Answer: 16
Explanation:
$ \therefore $ Mass of acrolein = $\left[ {{x \over {74}} \times 0.64} \right] \times 56 = 78$
$ \Rightarrow x = 16.1gm \simeq 16gm$
2021
Q167
JEE Mains
Numerical
14 Mar 2026
A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n $\times$ 10$-$1, when n = __________. (Round off to the Nearest Integer).
(Given : Atomic masses : C : 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]
(Given : Atomic masses : C : 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]
Correct Answer: 3
Explanation:
Number of moles of benzyl trimethyl
ammonium bromide formed = ${{23} \over {230}}$ = 0.1
$ \therefore $ No. of moles of bromomethane consumed
= 3 $ \times $ 0.1
= 3 $ \times $ 10–1
ammonium bromide formed = ${{23} \over {230}}$ = 0.1
$ \therefore $ No. of moles of bromomethane consumed
= 3 $ \times $ 0.1
= 3 $ \times $ 10–1
2021
Q168
JEE Mains
Numerical
14 Mar 2026
Complete combustion of 3g of ethane gives x $\times$ 1022 molecules of water. The value of x is __________. (Round off to the Nearest Integer). [Use : NA = 6.023 $\times$ 1023; Atomic masses in u : C : 12.0; O : 16.0; H : 1.0]
Correct Answer: 18
Explanation:
${C_2}{H_6} + {7 \over 2}{O_2}\buildrel {} \over
\longrightarrow 2C{O_{\,2}} + 3{H_2}O$
Here, ${C_2}{H_6} = 3gm = {3 \over {30}} = 0.1$ mol
From 1 mol C2H6 we get 3 mol H2O
$ \therefore $ From 0.1 mol C2H6 we get = 3 $\times$ 0.1 mol H2O
$ \therefore $ Moles of H2O = 0.3
= 0.3 $\times$ 6.023 $\times$ 1023 molecules
= 1.8069 $\times$ 1023 molecules
= 18.069 $\times$ 1022 molecules
Here, ${C_2}{H_6} = 3gm = {3 \over {30}} = 0.1$ mol
From 1 mol C2H6 we get 3 mol H2O
$ \therefore $ From 0.1 mol C2H6 we get = 3 $\times$ 0.1 mol H2O
$ \therefore $ Moles of H2O = 0.3
= 0.3 $\times$ 6.023 $\times$ 1023 molecules
= 1.8069 $\times$ 1023 molecules
= 18.069 $\times$ 1022 molecules
2021
Q169
JEE Mains
Numerical
14 Mar 2026
Consider the above reaction. The percentage yield of amide product is __________. (Round off to the Nearest Integer).
(Given : Atomic mass : C : 12.0 u, H : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
Correct Answer: 77
Explanation:
Stoichiometric moles of amide = 10$-$3 mol
Actual weight of amide = 10-3 $ \times $ 273 = 0.273 g
% yield = ${{0.210} \over {0.273}} \times 100$
= 76.9%
$ \simeq $ 77%
2021
Q170
JEE Mains
Numerical
14 Mar 2026
15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr2O$_7^{2 - }$. The molarity of the Fe2+ solution is __________ $\times$ 10-2 M. (Round off to the Nearest Integer).
Correct Answer: 24
Explanation:
$C{r_2}{O_7}^{2 - } + F{e^{2 + }}\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} C{r^{3 + }} + F{e^{3 + }}$
Valance Factor of $C{r_2}{O_7}^{2 - }$ = 6
Valance Factor of $F{e^{2 + }}$ = 1
mili eq. of $C{r_2}{O_7}^{2 - }$ = mili eq. of Fe2+
6(0.03 $\times$ 20) = 1(M $\times$ 15)
M = 6(0.03)4/3
M = 0.24 M
M = 24 $\times$ 10$-$2 M
Valance Factor of $C{r_2}{O_7}^{2 - }$ = 6
Valance Factor of $F{e^{2 + }}$ = 1
mili eq. of $C{r_2}{O_7}^{2 - }$ = mili eq. of Fe2+
6(0.03 $\times$ 20) = 1(M $\times$ 15)
M = 6(0.03)4/3
M = 0.24 M
M = 24 $\times$ 10$-$2 M
2021
Q171
JEE Mains
Numerical
14 Mar 2026
The mole fraction of a solute in a 100 molal aqueous solution is ___________ $\times$ 10$-$2. (Round off to the Nearest Integer).
[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
Correct Answer: 64
Explanation:
Let weight of H2O = 1000 g
Moles of solute = 100
(mole)H2O = ${{1000} \over {18}}$
Mole fraction of solute = ${{mole\,of\,solute} \over {Total\,moles}}$
$ = {{100} \over {100 + {{1000} \over {18}}}} = {{1800} \over {2800}}$
${X_{solute}} = 64 \times {10^{ - 2}}$
Moles of solute = 100
(mole)H2O = ${{1000} \over {18}}$
Mole fraction of solute = ${{mole\,of\,solute} \over {Total\,moles}}$
$ = {{100} \over {100 + {{1000} \over {18}}}} = {{1800} \over {2800}}$
${X_{solute}} = 64 \times {10^{ - 2}}$
2021
Q172
JEE Mains
Numerical
14 Mar 2026
When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, _________ $\times$ 10$-$5 moles of lead sulphate precipitate out. (Round off to the Nearest Integer).
Correct Answer: 525
Explanation:
For 3 moles of Pb(NO3)2 , we require 1 mole of Cr2(SO4)3
For 5.25 moles of Pb(NO3)2, we require $\frac{1}{3} \times 5.25 $ mole of Cr2(SO4)3 = 1.75 moles
But we have 2.4 moles. So, Cr2(SO4)3 is excess reagent and Pb(NO3)2 is limiting reagent, (LR)
Moles of PbSO4 formed = moles of Pb(NO3)2 consumed
= 5.25 m mol = 525 $ \times $ 10-5 moles
2021
Q173
JEE Mains
Numerical
14 Mar 2026
Complete combustion of 750 g of an organic compound provides 420 g of CO2 and 210 g of H2O. The percentage composition of carbon and hydrogen in organic compound is 15.3 and ___________ respectively. (Round off to the Nearest Integer).
Correct Answer: 3
Explanation:
18 gm H2O $ \Rightarrow $ 2 gm H2
210 gm $ \Rightarrow $ ${2 \over {18}} \times 210$
= 23.33 gm H2
So, % H2 = ${{23.33} \over {750}} \times 100$ = 3.11% $ \approx $ 3%
210 gm $ \Rightarrow $ ${2 \over {18}} \times 210$
= 23.33 gm H2
So, % H2 = ${{23.33} \over {750}} \times 100$ = 3.11% $ \approx $ 3%
2021
Q174
JEE Mains
Numerical
14 Mar 2026
$2MnO_4^ - + b{C_2}O_4^{2 - } + c{H^ + } \to xM{n^{2 + }} + yC{O_2} + z{H_2}O$
If the above equation is balanced with integer coefficients, the value of c is ___________. (Round off to the Nearest Integer).
If the above equation is balanced with integer coefficients, the value of c is ___________. (Round off to the Nearest Integer).
Correct Answer: 16
Explanation:
$2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$
2021
Q175
JEE Mains
Numerical
14 Mar 2026
A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm$-$3. The molarity of the solution is ____________ mol dm$-$3. (Round off to the Nearest Integer).
[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]
[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]
Correct Answer: 9
Explanation:
$m = {{1000 \times M} \over {1000 \times d - M \times {M_{solute}}}}$
$6.5 = {{1000 \times M} \over {1890 - M \times 56}}$
$ \Rightarrow $ $12285 - 364M = 1000M$
$ \Rightarrow $ $1364M = 12285$
$ \Rightarrow $ $M = 9$
$6.5 = {{1000 \times M} \over {1890 - M \times 56}}$
$ \Rightarrow $ $12285 - 364M = 1000M$
$ \Rightarrow $ $1364M = 12285$
$ \Rightarrow $ $M = 9$
2021
Q176
JEE Mains
Numerical
14 Mar 2026
The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+ per mL is _________ g. (Rounded off to the nearest integer)
[Given : Atomic weight in g mol$-$1 - Na : 23; N : 14; O : 16]
[Given : Atomic weight in g mol$-$1 - Na : 23; N : 14; O : 16]
Correct Answer: 13
Explanation:
Na+ = 70 mg/mL
WNa+ in 50 mL solution
= 70 $\times$ 50 mg
= 3500 mg
= 3.5 gm
Moles of Na+ in 50 mL solution = ${{3.5} \over {23}}$
Moles of NaNO3 = moles of Na+
= ${{3.5} \over {23}}$ mol
Mass of NaNO3 = ${{3.5} \over {23}} \times 85 = 12.934$
$ \simeq $ 13 gm
WNa+ in 50 mL solution
= 70 $\times$ 50 mg
= 3500 mg
= 3.5 gm
Moles of Na+ in 50 mL solution = ${{3.5} \over {23}}$
Moles of NaNO3 = moles of Na+
= ${{3.5} \over {23}}$ mol
Mass of NaNO3 = ${{3.5} \over {23}} \times 85 = 12.934$
$ \simeq $ 13 gm
2021
Q177
JEE Mains
Numerical
14 Mar 2026
The number of significant figures in 50000.020 $\times$ 10$-$3 is _____________.
Correct Answer: 8
Explanation:
10$-$3 has no role in significant digits. Here in 50000.020, Number of significant figure = 8 as all zeroes between non zero digits are counted as significant digits and also the last zero is also significant digit as zeroes at the end or right of the number is significant only if they are present at the right side of the decimal.
2021
Q178
JEE Mains
Numerical
14 Mar 2026
In basic medium $Cr{O_4}^{2 - }$ oxidises ${S_2}{O_3}^{2 - }$ to form $S{O_4}^{2 - }$ and itself changes into $Cr{(OH)_4}^ - $. The volume of 0.154 M $Cr{O_4}^{2 - }$ required to react with 40 mL of 0.25 M ${S_2}{O_3}^{2 - }$ is __________ mL. (Rounded off to the nearest integer)
Correct Answer: 173
Explanation:
$17{H_2}O + 8Cr{O_4} + 3{S_2}{O_3}\buildrel {} \over
\longrightarrow 6S{O_4} + 8Cr{(OH)_4}^ - + 2O{H^ - }$
Applying mole-mole analysis
${{0.154 \times v} \over 8} = {{40 \times 0.25} \over 3}$
v = 173 mL
Applying mole-mole analysis
${{0.154 \times v} \over 8} = {{40 \times 0.25} \over 3}$
v = 173 mL
2021
Q179
JEE Mains
Numerical
14 Mar 2026
The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is _____________.
Correct Answer: 8
Explanation:
Combustion reaction :
${C_x}{H_y}(g) + \left( {x + {y \over 4}} \right){O_2}(g) \to xC{O_2}(g) + {y \over 2}{H_2}O(l)$
Suppose, volume of CxHy is V and volume of O2 is 6 times greater than CxHy = 6V
then volume of xCO2 $\Rightarrow$ Vx = 4 V
x = 4
Since, ${V_{{O_2}}} = 6 \times {V_{{C_x}{H_y}}}$
$V\left( {x + {y \over 4}} \right)$ = 6V
$\left( {x + {y \over 4}} \right) = 6$ ..... (i)
Put value of x = 4 in Eq. (i) we get,
$4 + {y \over 4} = 6 \Rightarrow y = 8$
${C_x}{H_y}(g) + \left( {x + {y \over 4}} \right){O_2}(g) \to xC{O_2}(g) + {y \over 2}{H_2}O(l)$
Suppose, volume of CxHy is V and volume of O2 is 6 times greater than CxHy = 6V
then volume of xCO2 $\Rightarrow$ Vx = 4 V
x = 4
Since, ${V_{{O_2}}} = 6 \times {V_{{C_x}{H_y}}}$
$V\left( {x + {y \over 4}} \right)$ = 6V
$\left( {x + {y \over 4}} \right) = 6$ ..... (i)
Put value of x = 4 in Eq. (i) we get,
$4 + {y \over 4} = 6 \Rightarrow y = 8$
2021
Q180
JEE Mains
Numerical
14 Mar 2026
1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is __________ $\times$ 10$-$2.
Correct Answer: 243
Explanation:
Given, weight = 18.6 g
Here, 1 mole of aniline gives 1 mole of acetanilide
$\therefore$ mole of aniline = mole of acetanilide
$\Rightarrow$ ${{1.86} \over {93}} = {{{W_\text{Acetanilide}}} \over {135}}$
${W_\text{Acetanilide}} = {{1.86 \times 135} \over {93}}g = 2.70g$
But efficiency of reaction is 90% only.
Hence, mass of acetanilide produced $ = 2.70 \times {{90} \over {100}}g = 2.43g = 243 \times {10^2}g$
x = 243
2021
Q181
JEE Mains
Numerical
14 Mar 2026
4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The
molarity of the solution in M is x $ \times $ 10-1.
The value of x is _______. (Rounded off to the nearest integer)
The value of x is _______. (Rounded off to the nearest integer)
Correct Answer: 2
Explanation:
Given, weight of compound A = 4.5 g
Molecular weight of compound A = 90 g/mol
Volume of solution (in mL) = 250 mL
Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).
$M = {{Number\,of\,moles\,of\,solute\,(n)} \over {Volume\,of\,solution}}$
$ = {{{{4.5} \over {90}}} \over {{{250} \over {1000}}}} = 0.2$ = 2 $\times$ 10$-$1 M
$\therefore$ $n = {{Weight\,of\,solute\,(compound\,A)} \over {Molecular\,weight\,of\,solute\,(compound\,A)}}$
Hence, x $\times$ 10$-$1 $\mu$
x = 2
Molecular weight of compound A = 90 g/mol
Volume of solution (in mL) = 250 mL
Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).
$M = {{Number\,of\,moles\,of\,solute\,(n)} \over {Volume\,of\,solution}}$
$ = {{{{4.5} \over {90}}} \over {{{250} \over {1000}}}} = 0.2$ = 2 $\times$ 10$-$1 M
$\therefore$ $n = {{Weight\,of\,solute\,(compound\,A)} \over {Molecular\,weight\,of\,solute\,(compound\,A)}}$
Hence, x $\times$ 10$-$1 $\mu$
x = 2
2020
Q182
JEE Mains
MCQ
14 Mar 2026
The average molar mass of chlorine is
35.5 g mol–1. The ratio of 35Cl to 37Cl in naturally
occuring chlorine is close to :
A.
1 : 1
B.
2 : 1
C.
3 : 1
D.
4 : 1
2020
Q183
JEE Mains
MCQ
14 Mar 2026
A solution of two components containing
n1 moles of the 1st component and n2 moles of
the 2nd component is prepared. M1 and M2 are
the molecular weights of component 1 and 2
respectively. If d is the density of the solution
in g mL–1, C2 is the molarity and x2 is the mole
fraction of the 2nd component, then C2 can be
expressed as :
A.
${C_2} = {{1000{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$
B.
${C_2} = {{1000d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$
C.
${C_2} = {{d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$
D.
${C_2} = {{d{x_1}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$
2020
Q184
JEE Mains
MCQ
14 Mar 2026
The strengths of 5.6 volume hydrogen peroxide
(of density 1 g/mL) in terms of mass percentage
and molarity (M), respectively, are:
(Take molar mass of hydrogen peroxide as
34 g/mol)
A.
0.85 and 0.5
B.
0.85 and 0.25
C.
1.7 and 0.25
D.
1.7 and 0.5
2020
Q185
JEE Mains
MCQ
14 Mar 2026
The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by :
A.
200 ml of 0.02 N HCl
B.
100 ml of 0.2 N HCl
C.
100 ml of 0.1 HCl
D.
200 ml of 0.4 N HCl
2020
Q186
JEE Mains
MCQ
14 Mar 2026
Amongst the following statements, that which was not proposed by Dalton was :
A.
Matter consists of indivisible atoms all the atoms of a given element have.
B.
Chemical reactions involve reorganization of atoms. These are neither created not destroyed in
a chemical reaction.
C.
When gases combine or reproduced in a chemical reactionn they do so in a simple ratio by
volume provided all gases are the same T & P.
D.
Identical properties including identical mass. Atoms of differemt element differ in mass.
2020
Q187
JEE Mains
Numerical
14 Mar 2026
Consider the following equations :
2Fe2+ + H2O2 $ \to $ xA + yB
(in basic medium)
2MnO4- + 6H+ + 5H2O2 $ \to $ x'C + y'D + z'E
(in acidic medium)
The sum of the stoichiometric coefficients x, y, x', y', and z' for products A, B, C, D and E, respectively, is ______.
2Fe2+ + H2O2 $ \to $ xA + yB
(in basic medium)
2MnO4- + 6H+ + 5H2O2 $ \to $ x'C + y'D + z'E
(in acidic medium)
The sum of the stoichiometric coefficients x, y, x', y', and z' for products A, B, C, D and E, respectively, is ______.
Correct Answer: 19
Explanation:
2Fe2+ + H2O2 $ \to $ 2Fe3+ + 2OH–
2MnO4- + 6H+ + 5H2O2 $ \to $ 2Mn2+ + 8H2O + 5O2
$ \therefore $ x + y + x' + y' + z' = 19
2MnO4- + 6H+ + 5H2O2 $ \to $ 2Mn2+ + 8H2O + 5O2
$ \therefore $ x + y + x' + y' + z' = 19
2020
Q188
JEE Mains
Numerical
14 Mar 2026
A 20.0 mL solution containing 0.2 g impure
H2O2 reacts completely with 0.316 g of KMnO4
in acid solution. The purity of H2O2 (in %) is
_____________
(mol. wt. of H2O2 = 34; mol. wt. of KMnO4 = 158)
(mol. wt. of H2O2 = 34; mol. wt. of KMnO4 = 158)
Correct Answer: 85
Explanation:
5H2O2 + 2MnO4- + 6H+ $ \to $ 2Mn2+ + 5O2 + 8H2O
Moles of KMnO4 = ${{0.316} \over {158}}$ = 2 $ \times $ 10-3
Equivalents of H2O2 = Equivalent of KMnO4
= 2 × 10–3 × 5 = 0.01
Moles of H2O2 = ${{0.01} \over 2}$ = 0.005
Mass of pure H2O2 = 0.005 × 34 = 0.170 gm
Percentage purity = ${{0.17} \over {0.2}}$ $ \times $ 100 = 85 %
Moles of KMnO4 = ${{0.316} \over {158}}$ = 2 $ \times $ 10-3
Equivalents of H2O2 = Equivalent of KMnO4
= 2 × 10–3 × 5 = 0.01
Moles of H2O2 = ${{0.01} \over 2}$ = 0.005
Mass of pure H2O2 = 0.005 × 34 = 0.170 gm
Percentage purity = ${{0.17} \over {0.2}}$ $ \times $ 100 = 85 %
2020
Q189
JEE Mains
Numerical
14 Mar 2026
The mass of ammonia in grams produced when
2.8 kg of dinitrogen quantitatively reacts with 1
kg of dihydrogen is _______.
Correct Answer: 3400
Explanation:
N2(g) + 3H2(g) $ \to $ 2NH3(g)
Number of moles of N2 = ${{2.8 \times {{10}^3}} \over {28}}$ = 100
Number of moles of H2 = ${{1000} \over 2}$ = 500
Here N2 is limiting reagent.
$ \therefore $ Number of moles of NH3 produced = 2 $ \times $ 100 = 200
Mass of NH3 produced = 200 × 17 = 3400 gm
Number of moles of N2 = ${{2.8 \times {{10}^3}} \over {28}}$ = 100
Number of moles of H2 = ${{1000} \over 2}$ = 500
Here N2 is limiting reagent.
$ \therefore $ Number of moles of NH3 produced = 2 $ \times $ 100 = 200
Mass of NH3 produced = 200 × 17 = 3400 gm
2020
Q190
JEE Mains
Numerical
14 Mar 2026
6.023 $ \times $ 1022 molecules are present in 10 g of a substance 'x'. The molarity of a solution containing
5 g of substance 'x' in 2 L solution is _____ × 10-3
Correct Answer: 25
Explanation:
Mass of 6.023 × 1022 molecules of a substance
= 10 g
Mass of 6.023 × 1023 molecules of the substance = 100 g
$ \therefore $ Molar mass of the substance = 100 g mol–1
Molarity of the solution =
= ${{\left( {5/100} \right)} \over 2}$ = 0.025
Mass of 6.023 × 1023 molecules of the substance = 100 g
$ \therefore $ Molar mass of the substance = 100 g mol–1
Molarity of the solution =
moles of solute
volume of solution(in l)
= ${{\left( {5/100} \right)} \over 2}$ = 0.025
2020
Q191
JEE Mains
Numerical
14 Mar 2026
The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ___________.
Correct Answer: 10
Explanation:
H3PO2 + NaOH $ \to $ NaH2PO2 + H2O
Using Stoichiometry
$ \Rightarrow $ ${{0.1 \times 10} \over 1}$ = 0.1 × VNaOH
$ \Rightarrow $ VNaOH = 10 ml
Using Stoichiometry
Moles of H3PO2 reacted
1
=
Moles of NaOH reacted
1
$ \Rightarrow $ ${{0.1 \times 10} \over 1}$ = 0.1 × VNaOH
$ \Rightarrow $ VNaOH = 10 ml
2020
Q192
JEE Mains
Numerical
14 Mar 2026
The mole fraction of glucose (C6H12O6
) in an aqueous binary solution is 0.1. The mass percentage of
water in it, to the nearest integer, is _______.
Correct Answer: 47
Explanation:
Mole fraction of glucose in aqueous solution
= 0.1
Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9.
So mass % of water = ${{0.9 \times 18} \over {0.1 \times 180 + 0.9 \times 18}}$ $ \times $ 100
= 47.37 $ \simeq $ 47
Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9.
So mass % of water = ${{0.9 \times 18} \over {0.1 \times 180 + 0.9 \times 18}}$ $ \times $ 100
= 47.37 $ \simeq $ 47
2020
Q193
JEE Mains
Numerical
14 Mar 2026
The volume strength of 8.9 M H2O2
solution calculated at 273 K and 1 atm is ______. (R = 0.0821 L
atm K-1 mol-1) (rounded off ot the nearest integer)
Correct Answer: 100
Explanation:
Volume strength of H2O2 at 1 atm
273 kelvin
= M × 11.2 = 8.9 × 11.2 = 99.68 $ \simeq $ 100
= M × 11.2 = 8.9 × 11.2 = 99.68 $ \simeq $ 100
2020
Q194
JEE Mains
Numerical
14 Mar 2026
The ratio of the mass percentages of ‘C & H’
and ‘C & O’ of a saturated acyclic organic
compound ‘X’ are 4 : 1 and 3 : 4 respectively.
Then, the moles of oxygen gas required for
complete combustion of two moles of organic
compound ‘X’ is ________.
Correct Answer: 5
Explanation:
Let the organic compound X is = CxHyOz
Here moles of C = x, moles of H = y, moles of O = z
Given, ${{{W_C}} \over {{W_H}}} = {4 \over 1}$
$ \therefore $ ${x \over y} = {{{{{W_C}} \over {12}}} \over {{{{W_H}} \over 1}}}$ = ${4 \over 1} \times {1 \over {12}}$ = ${1 \over 3}$
Also Given, ${{{W_C}} \over {{W_O}}} = {3 \over 4}$
$ \therefore $ ${x \over z} = {{{{{W_C}} \over {12}}} \over {{{{W_O}} \over {16}}}}$ = ${3 \over 4} \times {{16} \over {12}}$ = 1
$ \Rightarrow $ x = z
$ \therefore $ Empirical formula = CxH3xOx = CH3O
As compound is saturated acyclic organic
compound, so molecular formula = C2H6O2
C2H6O2 + ${5 \over 2}$O2 $ \to $ 2CO2 + 3H2O
For 1 mole of C2H6O2 number of moles of O2 required = ${5 \over 2}$
$ \therefore $ For 2 mole of C2H6O2 number of moles
of O2 required = ${5 \over 2}$ $ \times $ 2 = 5
Here moles of C = x, moles of H = y, moles of O = z
Given, ${{{W_C}} \over {{W_H}}} = {4 \over 1}$
$ \therefore $ ${x \over y} = {{{{{W_C}} \over {12}}} \over {{{{W_H}} \over 1}}}$ = ${4 \over 1} \times {1 \over {12}}$ = ${1 \over 3}$
Also Given, ${{{W_C}} \over {{W_O}}} = {3 \over 4}$
$ \therefore $ ${x \over z} = {{{{{W_C}} \over {12}}} \over {{{{W_O}} \over {16}}}}$ = ${3 \over 4} \times {{16} \over {12}}$ = 1
$ \Rightarrow $ x = z
$ \therefore $ Empirical formula = CxH3xOx = CH3O
As compound is saturated acyclic organic
compound, so molecular formula = C2H6O2
C2H6O2 + ${5 \over 2}$O2 $ \to $ 2CO2 + 3H2O
For 1 mole of C2H6O2 number of moles of O2 required = ${5 \over 2}$
$ \therefore $ For 2 mole of C2H6O2 number of moles
of O2 required = ${5 \over 2}$ $ \times $ 2 = 5
2020
Q195
JEE Mains
Numerical
14 Mar 2026
10.30 mg of O2 is dissolved into a liter of sea
water of density 1.03 g/mL. The concentration
of O2 in ppm is__________.
Correct Answer: 10
Explanation:
1030 gm of sea water contains = 10.3 × 10–3 gm
106 gm of sea water contains = ${{10.3 \times {{10}^{ - 3}}} \over {1030}} \times {10^6}$ = 10 ppm
106 gm of sea water contains = ${{10.3 \times {{10}^{ - 3}}} \over {1030}} \times {10^6}$ = 10 ppm
2020
Q196
JEE Mains
Numerical
14 Mar 2026
The molarity of HNO3 in a sample which has
density 1.4 g/mL and mass percentage of 63%
is _____.
(Molecular Weight of HNO3 = 63)
(Molecular Weight of HNO3 = 63)
Correct Answer: 14
Explanation:
%w/w = 63%
dsolution = 1.4 g/ml
Molarity =
= ${{63 \times 1.4 \times 10} \over {63}}$ = 14 M
dsolution = 1.4 g/ml
Molarity =
%w/w $ \times $ dsolution
Moleculer weight of Solute
.$ \times $ 10
= ${{63 \times 1.4 \times 10} \over {63}}$ = 14 M
2020
Q197
JEE Mains
Numerical
14 Mar 2026
The hardness of a water sample containing
10–3 M MgSO4 expressed as CaCO3 equivalents
(in ppm) is ______.
(molar mass of MgSO4 is 120.37 g/mol)
(molar mass of MgSO4 is 120.37 g/mol)
Correct Answer: 100
Explanation:
Equivalance of MgSO4 = Equivalance of CaCO3
$ \Rightarrow $ 10-3 $ \times $ 2 = [CaCO3] $ \times $ 2
$ \Rightarrow $ [CaCO3] = 10-3 M = 10-3 mol/lit
= 10-3 $ \times $ 100 g/lit
= 100 mg/lit
= 100 ppm
$ \Rightarrow $ 10-3 $ \times $ 2 = [CaCO3] $ \times $ 2
$ \Rightarrow $ [CaCO3] = 10-3 M = 10-3 mol/lit
= 10-3 $ \times $ 100 g/lit
= 100 mg/lit
= 100 ppm
2020
Q198
JEE Mains
Numerical
14 Mar 2026
The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of
[Co(NH3)6]Cl3 is ________.
M[Co(NH3)6Cl3] = 267.46 g/mol
MAgNO3 = 169.87 g/mol
M[Co(NH3)6Cl3] = 267.46 g/mol
MAgNO3 = 169.87 g/mol
Correct Answer: 26.80to27.00
Explanation:
[Co(NH3)6]Cl3 + 3AgNO3 $ \to $ 3AgCl
$ \Rightarrow $ ${{0.3} \over {267.46}} = {{0.125 \times v \times {{10}^{ - 3}}} \over 3}$
$ \Rightarrow $ v = 26.92 ml
Mole of [Co(NH3)6]Cl3
1
=
Mole of AgNO3
3
$ \Rightarrow $ ${{0.3} \over {267.46}} = {{0.125 \times v \times {{10}^{ - 3}}} \over 3}$
$ \Rightarrow $ v = 26.92 ml
2020
Q199
JEE Mains
Numerical
14 Mar 2026
Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to
achieve 10 ppm of iron in 100 kg of wheat is _______.
Atomic weight : Fe = 55.85; S = 32.00; O = 16.00
Atomic weight : Fe = 55.85; S = 32.00; O = 16.00
Correct Answer: 4.95to4.97
Explanation:
FeSO4.7H2O (M = 277.85)
PPM =
$ \Rightarrow $ 10 =
$ \Rightarrow $ Mass of Iron = 1 gm
Molecular mass of FeSO4.7H2O is 277.85
55.85 gm iron is present in 277.85 gm of salt
1 gm iron is present in = ${{277.85} \over {55.85}}$ = 4.97 gm of salt.
PPM =
Mass of Iron
Mass of wheat
$ \times $ 106
$ \Rightarrow $ 10 =
Mass of Iron
100 $ \times $ 103
$ \times $ 106
$ \Rightarrow $ Mass of Iron = 1 gm
Molecular mass of FeSO4.7H2O is 277.85
55.85 gm iron is present in 277.85 gm of salt
1 gm iron is present in = ${{277.85} \over {55.85}}$ = 4.97 gm of salt.
2020
Q200
JEE Mains
Numerical
14 Mar 2026
The flocculation value of HCl for arsenic sulphide sol.
is 30 m mol L-1 If H2SO4 is used for the flocculatiopn of arsenic sulphide, the amount in grams, of H2SO4 in 250 ml required for the above purposed is ______.
(molecular mass of H2SO4 = 98 g/mol)
is 30 m mol L-1 If H2SO4 is used for the flocculatiopn of arsenic sulphide, the amount in grams, of H2SO4 in 250 ml required for the above purposed is ______.
(molecular mass of H2SO4 = 98 g/mol)
Correct Answer: 0.36to0.38
Explanation:
Arsenic sulphide sol is negatively charged, so for flocculation
positive ion required. Here positive ion H+ present.
for 1 L, 30 mm moles of H+ is required
for 250 ml, ${{30} \over 4}$ mm moles H+ is required
$ \therefore $ for 250 ml, ${{30} \over {4 \times 2}}$ mm moles H2SO4 is required.
$ \therefore $ Weight of H2SO4 required = ${{30} \over {4 \times 2}} \times {10^{ - 3}} \times 98$
= 0.3675 g
for 1 L, 30 mm moles of H+ is required
for 250 ml, ${{30} \over 4}$ mm moles H+ is required
$ \therefore $ for 250 ml, ${{30} \over {4 \times 2}}$ mm moles H2SO4 is required.
$ \therefore $ Weight of H2SO4 required = ${{30} \over {4 \times 2}} \times {10^{ - 3}} \times 98$
= 0.3675 g