Some Basic Concepts of Chemistry
348 Questions
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
$0.63 \mathrm{~g}$ of oxalic acid is dissolved in order to obtain $250 \mathrm{~cm}^3$ of its solution. Find the normality of this solution. [oxalic acid $\left.(\mathrm{COOH})_2 \cdot 2 \mathrm{H}_2 \mathrm{O}\right]$
A.
0.05 N
B.
0.01 N
C.
0.04 N
D.
0.02 N
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
$7.8 \mathrm{~g}$ of a compound having molecular formula $\mathrm{C}_6 \mathrm{H}_6$, on reacting with $\mathrm{CH}_3 \mathrm{COCl} / \mathrm{AlCl}_3$ gives $8.4 \mathrm{~g}$ of a product which has molecular formula $\mathrm{C}_8 \mathrm{H}_8 \mathrm{O}$. Calculate the percentage yield of the product $\mathrm{C}_8 \mathrm{H}_8 \mathrm{O}$. (Given, atomic weights of $\mathrm{H}, \mathrm{C}$ and $\mathrm{O}$ respectively are 1, 12 and 16)
A.
70%
B.
60%
C.
80%
D.
75%
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
The complete combustion of one mole of benzene produces .......... grams of carbon dioxide.
A.
164
B.
220
C.
264
D.
308
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Morning Shift
An alloy of metals X and Y weighs 12 g and contains atoms X and Y in the ratio of 2 : 5. The percentage of metal X in the alloy is 20 by mass. If the atomic mass of X is 40 amu what is the atomic mass of metal Y ?
A.
64 amu
B.
32 amu
C.
60 amu
D.
50 amu
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Evening Shift
If 0.2 moles of sulphuric acid is poured into 250 mL of water, calculate the concentration of the solution.
A.
0.8 N
B.
0.8 M
C.
8 M
D.
0.2 N
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Morning Shift
When 20 g of CaCO$_3$ is treated with 20 g of HCl, the mass of CO$_2$ formed would be
A.
10 g
B.
8.8 g
C.
22.2 g
D.
20 g
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
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
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Evening Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
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
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
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
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
In the chemical reaction between stoichiometric quantities of KMnO4 and KI in weakly basic solution, what is the number of moles of I2 released for 4 moles of KMnO4 consumed?
Correct Answer: 6
Explanation:
In alkaline medium : Iodide is oxidised to iodate
$2MnO_4^ - + {H_2}O + {I^ - }\buildrel {} \over \longrightarrow 2Mn{O_2} + 2O{H^ - } + IO_3^ - $
But in weakly basic solution :
$\mathop {KMn{O_4}}\limits^{ + 7} + \mathop {KI}\limits^{ - 1} \buildrel {} \over \longrightarrow \mathop {Mn{O_2}}\limits^{ + 4} + \mathop {{I_2}}\limits^0 $
Eq. of KMnO4 = Eq. of I2
$4 \times 3 = n \times 2 \Rightarrow n = 6$
$2MnO_4^ - + {H_2}O + {I^ - }\buildrel {} \over \longrightarrow 2Mn{O_2} + 2O{H^ - } + IO_3^ - $
But in weakly basic solution :
$\mathop {KMn{O_4}}\limits^{ + 7} + \mathop {KI}\limits^{ - 1} \buildrel {} \over \longrightarrow \mathop {Mn{O_2}}\limits^{ + 4} + \mathop {{I_2}}\limits^0 $
Eq. of KMnO4 = Eq. of I2
$4 \times 3 = n \times 2 \Rightarrow n = 6$
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 1 Offline
5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution?
| Exp. No. | Vol. of NaOH (mL) |
|---|---|
| 1 | 12.5 |
| 2 | 10.5 |
| 3 | 9.0 |
| 4 | 9.0 |
| 5 | 9.0 |
Correct Answer: 0.11
Explanation:
Oxalic acid solution titrated with NaOH solution using phenolphthalein as an indicator.
${H_2}{C_2}{O_4} + 2NaOH\buildrel {} \over \longrightarrow N{a_2}{C_2}{O_4} + 2{H_2}O$
Equivalent of H2C2O4 reacted = Equivalent of NaOH reacted
$ = {{5 \times 2 \times 0.1} \over {1000}} = {{9 \times {M_{(NaOH)}} \times 1} \over {1000}}$
${M_{(NaOH)}} = {1 \over 9} = 0.11$
${H_2}{C_2}{O_4} + 2NaOH\buildrel {} \over \longrightarrow N{a_2}{C_2}{O_4} + 2{H_2}O$
Equivalent of H2C2O4 reacted = Equivalent of NaOH reacted
$ = {{5 \times 2 \times 0.1} \over {1000}} = {{9 \times {M_{(NaOH)}} \times 1} \over {1000}}$
${M_{(NaOH)}} = {1 \over 9} = 0.11$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown
hydrocarbon contains :
A.
18 g of carbon and 7 g of hydrogen
B.
20 g of carbon and 5 g of hydrogen
C.
22 g of carbon and 3 g of hydrogen
D.
24 g of carbon and 1 g of hydrogen
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and gaseous product.
MnO2 reacts with NaCl and concentrated H2O4 to give a pungent gas Z. X, Y and Z, respectively, are :
A.
KMnO4, K2MnO4 and Cl2
B.
K2MnO4, KMnO4 and SO2
C.
K3MnO4, K2MnO4 and Cl2
D.
K2MnO4, KMnO4 and Cl2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg–1
) of the
aqueous solution is :
A.
13.88 × 10–1
B.
13.88 × 10–3
C.
13.88
D.
13.88 × 10–2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
5 moles of AB2 weigh 125 × 10–3
kg and 10 moles of A2B2 weigh 300 × 10–3
kg. The molar mass of A (MA)
and molar mass of B (MB) in kg mol are:
A.
MA = 10 × 10–3
and MB = 5 × 10–3
B.
MA = 25 × 10–3
and MB = 50 × 10–3
C.
MA = 5 × 10–3
and MB = 10 × 10–3
D.
MA = 50 × 10–3
and MB = 25 × 10–3
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The minimum amount of O2(g) consumed per gram of reactant is for the reaction :
(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
A.
4Fe(s) + 3O2(g) $ \to $ 2Fe2O3(s)
B.
P4(s) + 5O2(g) $ \to $ P4O10(s)
C.
C3H8(g) + 5O2(g) $ \to $ 3CO2(g) + 4H2O(l)
D.
2Mg(s) + O2(g) $ \to $ 2MgO(s)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete
combustion, and 40 mL of CO2 is formed. The formula of the hydrocarbon is :
A.
C4H7Cl
B.
C4H6
C.
C4H8
D.
C4H10
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
What would be the molality of 20% (mass/
mass) aqueous solution of KI?
(molar mass of KI = 166 g mol–1)
(molar mass of KI = 166 g mol–1)
A.
1.51
B.
1.35
C.
1.08
D.
1.48
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
For a reaction,
N2(g) + 3H2(g) $ \to $ 2NH3(g) ;
identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures.
N2(g) + 3H2(g) $ \to $ 2NH3(g) ;
identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures.
A.
56g of N2 + 10g of H2
B.
14g of N2 + 4g of H2
C.
28g of N2 + 6g of H2
D.
35g of N2 + 8g of H2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
The percentage composition of carbon by mole
in methane is :
A.
80%
B.
20%
C.
75%
D.
25%
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
8 g of NaOH is dissolved in 18g of H2O. Mole fraction of NaOH in solution and molality (in mol kg–1) of the solution respectively are -
A.
0.2, 11.11
B.
0.167, 22.20
C.
0.167, 11.11
D.
0.2, 22.20
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ?
A.
50 mL
B.
12.5 mL
C.
25 mL
D.
75 mL
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
An organic compound is estimated through Duma's method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is :
A.
C6H8N
B.
C6H8N2
C.
C12H8N
D.
C12H8N2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol–1]
[Molar mass of NaHCO3 = 84 g mol–1]
A.
33.6
B.
0.84
C.
8.4
D.
16.8
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
The amount of sugar (C12H22O11) required to prepare 2L of its 0.1 M aqueous solution is :
A.
17.1 g
B.
34.2 g
C.
68.4 g
D.
136.8 g
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
For the following reaction, in the mass of water produced from 445 g of C57H110O6 is :
2C57H110O6(s) + 163 O2(g) $ \to $ 114 CO2(g) + 110 H2O(l)
2C57H110O6(s) + 163 O2(g) $ \to $ 114 CO2(g) + 110 H2O(l)
A.
490 g
B.
445 g
C.
495 g
D.
890 g
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg$-$1 is :
A.
12
B.
4
C.
8
D.
16
2019
JEE Advanced
Numerical
JEE Advanced 2019 Paper 2 Offline
The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm$-$3, then molarity of urea solution is ................
(Given data : Molar masses of urea and water are 60 g mol$-$1 and 18 g mol$-$1, respectively)
(Given data : Molar masses of urea and water are 60 g mol$-$1 and 18 g mol$-$1, respectively)
Correct Answer: 2.98
Explanation:
Molarity
$(M) = {{Number\,of\,moles\,of\,solute \times 1000} \over {Volume\,of\,solution\,(in\,mL)}}$
Also, volume = ${{Mass} \over {Density}}$
Given, mole fraction of urea $({\chi _{urea}})$ = 0.05
Mass of water = 900 g
Density = 1.2 g/cm3
${\chi _{urea}}$ = ${{{n_{urea}}} \over {{n_{urea}} + 50}}$
[$ \because $ Moles of water = ${{900} \over {18}}$ = 50]
$0.05 = $${{{n_{urea}}} \over {{n_{urea}} + 50}}$
$ \Rightarrow $ $19{n_{urea}} = 50$
${n_{urea}} = 2.6315$ moles
${w_{urea}} = {n_{urea}} \times {(M.wt)_{urea}}$
$ = (2.6315 \times 60)g$
$V = {{2.6315 \times 60 + 900} \over {1.2}}$
[$ \because $ $Density = {{Mass\,of\,solution} \over {Volume\,of\,solution}}$]
= 881.57 mL
Now, molarity
= $Number\,of\,moles\,of\,solute \times {{1000} \over {Volume\,of\,solution\,(mL)}}$
$ = {{2.6315 \times 1000} \over {881.57}}$ = 2.98 M
$(M) = {{Number\,of\,moles\,of\,solute \times 1000} \over {Volume\,of\,solution\,(in\,mL)}}$
Also, volume = ${{Mass} \over {Density}}$
Given, mole fraction of urea $({\chi _{urea}})$ = 0.05
Mass of water = 900 g
Density = 1.2 g/cm3
${\chi _{urea}}$ = ${{{n_{urea}}} \over {{n_{urea}} + 50}}$
[$ \because $ Moles of water = ${{900} \over {18}}$ = 50]
$0.05 = $${{{n_{urea}}} \over {{n_{urea}} + 50}}$
$ \Rightarrow $ $19{n_{urea}} = 50$
${n_{urea}} = 2.6315$ moles
${w_{urea}} = {n_{urea}} \times {(M.wt)_{urea}}$
$ = (2.6315 \times 60)g$
$V = {{2.6315 \times 60 + 900} \over {1.2}}$
[$ \because $ $Density = {{Mass\,of\,solution} \over {Volume\,of\,solution}}$]
= 881.57 mL
Now, molarity
= $Number\,of\,moles\,of\,solute \times {{1000} \over {Volume\,of\,solution\,(mL)}}$
$ = {{2.6315 \times 1000} \over {881.57}}$ = 2.98 M
2019
JEE Advanced
Numerical
JEE Advanced 2019 Paper 2 Offline
The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc. HNO3 to a compound with the highest oxidation state of sulphur is ..............
(Given data : Molar mass of water = 18 g mol$-$1)
(Given data : Molar mass of water = 18 g mol$-$1)
Correct Answer: 288
Explanation:
When rhombic sulphur (S8) is oxidised by conc. HNO3 then H2SO4 is obtained and NO2 gas is released.
${S_8} + 48HN{O_3}\buildrel {} \over \longrightarrow 8{H_2}S{O_4} + 48N{O_2} + 16{H_2}O$
1 mole of rhombic sulphur produces = 16 moles of H2O
$ \therefore $ Mass of water = 16 $ \times $ 18 (molar mass of H2O) = 288 g
${S_8} + 48HN{O_3}\buildrel {} \over \longrightarrow 8{H_2}S{O_4} + 48N{O_2} + 16{H_2}O$
1 mole of rhombic sulphur produces = 16 moles of H2O
$ \therefore $ Mass of water = 16 $ \times $ 18 (molar mass of H2O) = 288 g
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $ \times $ 1023 mol-1)
(Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $ \times $ 1023 mol-1)
A.
6.023 $ \times $ 1020
B.
6.023 $ \times $ 109
C.
6.023 $ \times $ 1021
D.
6.023 $ \times $ 1023
2018
JEE Mains
MCQ
JEE Main 2018 (Offline)
The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is
A.
C2H4O3
B.
C3H6O3
C.
C2H4O
D.
C3H4O2
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Evening Slot
For per gram of reactant, the maximum quantity of N2 gas is produced in which of the following thermal decomposition reactions ?
(Given : Atomic wt. - Cr = 52 u, Ba = 137 u)
(Given : Atomic wt. - Cr = 52 u, Ba = 137 u)
A.
(NH4)2Cr2O7(s) $ \to $ N2(g) + 4H2O(g) + Cr2O3(s)
B.
2NH4NO3(s) $ \to $ 2 N2(g) + 4H2O(g) + O2(g)
C.
Ba(N3)2(s) $ \to $ Ba(s) + 3N2(g)
D.
2NH3(g) $ \to $ N2(g) + 3H2(g)
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Morning Slot
A sample of $NaCl{O_3}$ is converted by heat to $NaCl$ with a loss of $0.16$ $g$ of oxygen. The residue is dissolved in water and precipitated as $AgCl.$ The mass of $AgCl$ (in $g$) obtained will be : (Given : Molar mass of $AgCl=143.5$ $g$ $mo{l^{ - 1}}$)
A.
$0.35$
B.
$0.41$
C.
$0.48$
D.
$0.54$
2018
JEE Advanced
Numerical
JEE Advanced 2018 Paper 2 Offline
To measure the quantity of $MnC{l_2}$ dissolved in an aqueous solution, it was completely converted to $KMn{O_4}$ using the reaction,
$MnC{l_2} + {K_2}{S_2}{O_8} + {H_2}O \to KMn{O_4} + {H_2}S{O_4} + HCl$ (equation not balanced).
Few drops of concentrated $HCl$ were added to this solution and gently warmed. Further, oxalic acid ($225$ $mg$) was added in portions till the colour of the permanganate ion disappeared. The quantity of $MnC{l_2}$ (in mg) present in the initial solution is ____________.
(Atomic weights in $g\,\,mo{l^{ - 1}}:Mn = 55,Cl = 35.5$ )
$MnC{l_2} + {K_2}{S_2}{O_8} + {H_2}O \to KMn{O_4} + {H_2}S{O_4} + HCl$ (equation not balanced).
Few drops of concentrated $HCl$ were added to this solution and gently warmed. Further, oxalic acid ($225$ $mg$) was added in portions till the colour of the permanganate ion disappeared. The quantity of $MnC{l_2}$ (in mg) present in the initial solution is ____________.
(Atomic weights in $g\,\,mo{l^{ - 1}}:Mn = 55,Cl = 35.5$ )
Correct Answer: 126
Explanation:
Also,
$2MnO_4^ - + 5{C_2}O_4^{2 - } \to 2M{n^{2 + }} + 10C{O_2}$
Hence, $2M{n^{2 + }} \equiv 5{C_2}O_4^{2 - }$
$1MnC{l_2} \equiv 2.5{H_2}{C_2}{O_4}$
Oxalic acid taken = 225 mg
$ = {{225} \over {90}} = 2.5$ millimoles
Hence, MnCl2 = 1 millimole
= (55 + 71) = 126 mg