Assertion : (A) One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.
Reason : (R) Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard
it is the atomic mass expressed in gram
it is also called gram atom
one gram atom of an element contain `6xx10^(23) ` atoms
None of the above
`38.5`
`35.5`
`36`
None of these
Both the statements are true and Statement II is the correct explanation of Statement I
Both the statements are true but Statement II is not the correct explanation of Statement I.
Statement I is true, but Statement II is false
Statement I is false, but Statement II is true.
`1/12 ` of `C^(12)``
`1/14` of `O^(16)`
`1 g` of `H_2`
`1.66xx10^(-23) kg`
mole ratio
molecularity
atomicity
Avogadro's number
`M/2`
`M/4`
`M/7`
`M/5`
`M`
`M/5`
`M/6`
`M/4`
`26.7`
`8.9`
`26.89`
`17.8`
`158`
`79`
`52.66`
`31.6`
`45`
`90`
`126`
`63`
contains five electrons in its valence orbit
contains half filled p -orbitals
is a diatomic molecule
has variable valency
`9`
`27`
`18`
`36`
its molecular weight
half of its molecular weight
twice of its molecular weight
four times its molecular weight
`16`
`64`
`32`
`8`
`58.8`
`14.7`
`29.4`
`88.2`
I and II
II and Ill
I and Ill
All of these
Both the statements are true and Statement II is the correct explanation of Statement I
Both the statements are true but Statement II is not the correct explanation of Statement I.
Statement I is true, but Statement II is false
Statement I is false, but Statement II is true.
`38.5`
`74.4`
`35.5`
`71`
`14/(6.022 xx 10^(22)) g`
`28/(6.022xx 10^(23)) g`
`1/(6.022xx10^(23)) g`
`14 u`
16g of CO
28 g of `N_2`
14 g of `N_2`
1.0g of `H_2`
`3xx6.02xx10^(23)`
`5xx6.022xx10^(23) `
`6xx6.02xx10^(23)`
`7xx6.02xx10^(23)`
(This question may have multiple correct answers)
`0.5 g` hydrogen
`4.0 g` sulphur
`7.0 g` nitrogen
`2.3 g` sodium
`16` g of `O_2`(g) and 4g of `H_2`
`16`g of `O_2` and `44` g of `CO_2`
`28` g of `N_2` and `32` g of `O_2`
`12` g of C(s) and `23` g of Na(s)
`40xx6.022xx10^(23)`
`32xx6.022xx10^(23)`
`(40xx6xx10^(23))/32`
`(32xx6xx10^(23))/40`
`HClO`
`HClO_2`
`HClO_3`
`HClO_4`
`5 M`
`50 M`
`0.005 M`
`0.5 M`
`6.023xx10^(23)`
`1/(9.08) xx 10^(31)`
`(6.023)/(9.108) xx 10^(54)`
`1/(9.108xx6.023) xx 10^(8)`
I, II and Ill
II, Ill and IV
I, Ill and IV
All of these
Column I | Column II | ||
---|---|---|---|
(A) | 88 g of `CO_2` | (1) | 0.2 mol |
(B) | `6.022 xx 10^23` molecules of `H_2O` | (2) | 2 mol |
(C) | 5.6 L of `O_2` at STP | (3) | 1 mol |
(D) | 96 g of `O_2` | (4) | `6.022 xx 10^23` molecules |
(E) | 1 mole of any gas | (5) | 3 mol |
`1 xx 10^(23)`
`1.5 xx 10^(23)`
`2 xx 10^(23)`
`6xx10^(23)`
`18 g`of water
`16 g` of `O_2`
`4.4 g` of `O_2`
`16 g` of `CH_4`
`3xx10^(-26) kg`
`3xx10^(-25) kg`
`1.5 xx 10^(-25) kg`
`2.5xx10^(-26) kg`
`12.044xx10^(20)` molecules
`6.022xx10^(23)` molecules
`1xx10^(23)` molecules
`12.044xx10^(23)` molecules
`71 g` of chlorine
`48 g` of magnesium
`127 g` of iodine
`4 g` of hydrogen
0.1 m
1 M
0.5 m
1 m
`0.25`
`0.50`
`1`
`2`
`(6.022xx10^(-23))/6`
`(6.02xx10^(25))/18`
`(6.022xx10^(22))/6`
None of these
`22.4 L`
`2.24 L`
`0.224 L`
`0.1 L`
`4 g` He
`46 g` Na
`0.40 g` Ca
`12 g` He
`100`
`200`
`300`
`400`
Assertion : (A) Combustion of 16 g of methane gives 18 g of water
Reason : (R) In the combustion of methane, water is one of the products.
(This question may have multiple correct answers)
`640 g`
`160 g`
`80 g`
`320 g`
Total mass of iron and oxygen in reactants= total mass of iron and oxygen in product therefore it follows law of conservation of mass
Total mass of reactants = total mass of product, therefore, law of multiple proportions is followed
Amount of `Fe_2O_ 3` can be increased by taking any one of the reactants (iron or oxygen) in excess
Amount of `Fe_2O_3` produced will decrease if the amount of any one of the reactants {iron or oxygen) is taken in excess
(This question may have multiple correct answers)
`1.5 M`
`1.66 M`
`0.017 M`
`1.59 M`
`4 molL^(-1)`
`20 molL^(-1)`
`0.2 molL^(-1)`
`2 molL^(-1)`
List I | List II | ||
---|---|---|---|
(A) | molarity | (1) | Gram formula weight of solute in one litre of solution. |
(B) | Molality | (2) | Number of gram equivalents of solute dissolved in one litre of solution |
(C) | Normality | (3) | Number of moles of solute dissolved in 1 kg of solvent . |
(D) | Formality | (4) | Number of moles of solute in one litre of solution |
`A →4 , B → 3, C → 2 , D → 1`
`A → 1 , B → 2 , C → 3 , D → 4`
`A → 4 , B → 2 , C → 3 , D → 1`
`A → 2 , B → 3 , C → 4 , D → 1`