Skip to main content
Back to SOME BASIC CONCEPTS OF CHEMISTRY

Your Progress

0 / 436 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 421-436 of 436 questions
421 Medium
421 of 436
Application +4 | -1
What volume of oxygen gas at NTP is necessary for complete combustion of 20 L20 \mathrm{~L} of propane measured at 0∘C0^{\circ} \mathrm{C} and 760 mm760 \mathrm{~mm} pressure?
View answer and explanation
Correct option: C
99.93 L99.93 \mathrm{~L}
Option C (99.93 L) is correct because it is the volume of oxygen gas calculated using the stoichiometry of the combustion reaction and the ideal gas law.
422 Easy
422 of 436
Application +4 | -1
If H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} ionises as H2SO4+2H2O→2H3O++SO42{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} + 2{\mathrm{H}_2}\mathrm{O} \to 2{\mathrm{H}_3}{\mathrm{O}^ + } + \mathrm{SO}_4^2, then total number of ions produced by 0.1  M  H2SO4\mathrm{0}\mathrm{.1}\;\mathrm{M}\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} will be
View answer and explanation
Correct option: D
1.8×10231.8 \times {10^{23}}
Option D (1.8×10231.8 \times {10^{23}}) is correct because 0.1 M H2SO4H_2SO_4 produces 0.3 M ions, which translates to 1.8×10231.8 \times {10^{23}} ions.
423 Hard
423 of 436
Application +4 | -1
The ratio of masses of oxygen and nitrogen of a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is
View answer and explanation
Correct option: B
0.313888889
Option B (07:32) is correct because the ratio of the number of molecules is equal to the ratio of the number of moles, which can be calculated from the given mass ratio and molar masses.
424 Medium
424 of 436
Analyzing +4 | -1
One litre N2,78litre  O2{\mathrm{N}_2},\frac{7}{8}\mathrm{litre}\;{\mathrm{O}_2} and 1 litre COCO are taken in a mixture under indentical conditions of P and T. The amount of gases present in mixture is given by:
View answer and explanation
Correct option: C
wN2=wO2=wCO{w_{{\mathrm{N}_2}}} = {w_{{\mathrm{O}_2}}} = {w_{\mathrm{CO}}}
Option D is correct because boron in H2C=CH−B⊕H3{H_2}C = CH - \mathop B\limits^ \oplus {H_3} has a positive charge and lacks a vacant orbital, preventing resonance.
425 Medium
425 of 436
Application +4 | -1
50 litres of water containing Ca(HCO3)2\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 when converted into soft water required 22.2 g Ca(OH)222.2 \mathrm{~g} \mathrm{~Ca}(\mathrm{OH})_2. Calculate the amount of Ca(HCO3)2\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 present per litre of hard water.
View answer and explanation
Correct option: A
0.972 g L−10.972 \mathrm{~g} \mathrm{~L}^{-1}
Option A (0.972 g L−10.972 \mathrm{~g} \mathrm{~L}^{-1}) is correct because it is the calculated amount of Ca(HCO3)2\mathrm{Ca(HCO_3)_2} per liter based on the reaction with Ca(OH)2\mathrm{Ca(OH)_2}.
426 Medium
426 of 436
Application +4 | -1
Weight of an atom of an element is 6.644×10−23  g6.644 \times {10^{ - 23}}\;g. What will be the number of g atom of that element in 40 kg?
View answer and explanation
Correct option: A
103{10^3}
Option A (103{10^3}) is correct because the number of gram atoms is calculated by dividing the total number of atoms by Avogadro's number.
427 Easy
427 of 436
Remembering +4 | -1
The molecular mass of CaSO4⋅2H2O\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O} is

(At. mass of Ca = 40, S = 32)
View answer and explanation
Correct option: C
172 u
Option C (172 u) is correct because the molecular mass is calculated by summing the atomic masses of all atoms in the formula.
428 Hard
428 of 436
Application +4 | -1
21.75g of MnO2Mn{O_2} on reaction with HCl forms 2.8 L of Cl2(g)C{l_2}\left( g \right) at STP, the percentage purity of MnO2Mn{O_2} is (atomic mass of Mn = 55u)
MnO2+4HCl→MnCl2+Cl2+2H2OMn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O
View answer and explanation
Correct option: D
0.5
Option D (0.5) is correct because the percentage purity is calculated by comparing the actual volume of Cl2 produced to the theoretical volume based on the amount of MnO2 used.
429 Medium
429 of 436
Analyzing +4 | -1
Cu2S→Cu2++SO32−\mathrm{Cu}_2 \mathrm{S} \rightarrow \mathrm{Cu}^{2+}+\mathrm{SO}_3^{2-}, if M\mathrm{M} is the molar mass of Cu2S\mathrm{Cu}_2 \mathrm{S}, then its equivalent mass is given as
View answer and explanation
Correct option: A
M8\frac{M}{8}
The correct answer is A (M/8) because the n-factor for Cu2S in the given reaction is 8, representing the total change in oxidation state.
430 Hard
430 of 436
Application +4 | -1
A gaseous alkane is exploded with oxygen.
The volume of O2\text{O}_2 for complete combustion to
the volume of CO2\text{CO}_2 formed is in 7:47:4 ratio. The
molecular formula of alkane is
View answer and explanation
Correct option: C
C2H6{{\text{C}}_{2}}{{\text{H}}_{6}}
Option C (C2H6{{\text{C}}_{2}}{{\text{H}}_{6}}) is correct because the ratio of O2 volume to CO2 volume matches the stoichiometry of the combustion reaction for ethane.
431 Easy
431 of 436
Application +4 | -1
Dopamine is a neurotransmitter, a molecule that serves to transmit message in the brain. The chemical formula of dopamine is C8H11O2 N.\mathrm{C}_8 \mathrm{H}_{11} \mathrm{O}_2 \mathrm{~N}. How many moles are there in 1 g1 \mathrm{~g} of dopamine?
View answer and explanation
Correct option: A
0.006540.00654
Option A (0.00654) is correct because it is calculated by dividing the given mass by the molar mass of dopamine.
432 Medium
432 of 436
Analyzing +4 | -1
Statement-I: 12 parts by mass of carbon in CO\mathrm{CO} and CO2\mathrm{CO}_2 molecules combine with 16 and 32 parts by mass of oxygen.

Statement-II: A given compound always contains exactly the same proportion of elements by weight.
View answer and explanation
Correct option: B
Both statement I and II are incorrect
Option B (Both statement I and II are incorrect) is correct because statement I violates the law of multiple proportions, and statement II is the law of definite proportions, but it's incorrectly stated.
433 Hard
433 of 436
Application +4 | -1
Boron has two isotopes boron-10 and boron-11 whose percentage abundances are 19.6%19.6 \% and 80.4%80.4 \% respectively. What is the average atomic mass of boron?
View answer and explanation
Correct option: A
10.804 amu
Option A (10.804 amu) is correct because the average atomic mass is calculated by summing the product of each isotope's mass and its fractional abundance.
434 Medium
434 of 436
Remembering +4 | -1
Often, while calculating, there is a need to convert units from one system to other. The method used to accomplish is called
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the process of converting units from one system to another can be referred to by all the mentioned names.
435 Medium
435 of 436
Application +4 | -1
Volume in ml\mathrm{ml} of 11.2 volume strength H2O2\mathrm{H}_2 \mathrm{O}_2 required to exactly neutralize 20ml20 \mathrm{ml} of 0.2M0.2 \mathrm{M} of KMnO4\mathrm{KMnO}_4 in acid medium is
View answer and explanation
Correct option: A
10ml10 \mathrm{ml}
Option A (10 ml) is correct because it is calculated by equating the milliequivalents of KMnO4 and H2O2 using the volume strength concept.
436 Medium
436 of 436
Application +4 | -1
How many moles of H2SO4{H_2}S{O_4} are present in 4.9 g H2SO4{H_2}S{O_4}
View answer and explanation
Correct option: B
0.05
Option B (0.05) is correct because the number of moles is calculated by dividing the given mass by the molar mass of H2SO4.
Revision notes and key takeaways

Review SOME BASIC CONCEPTS OF CHEMISTRY in your textbook or class notes, then use this set to check your understanding.

Showing 421-436 of 436 questions
0 /436
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.