Chemistry Previous Year Questions Of Concept of Atomic, Molecular and Equivalent Masses for NDA
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Previous Year Questions Of Concept of Atomic, Molecular and Equivalent Masses for NDA

Previous Year Questions Of Atomic, Molecular and Equivalent Masses for NDA

Q 2764312255

20g of a salt is dissolved in 180g of water What is the mass percentage of the salt in the inhibition ? .
NDA Paper 2 2017
(A)

5%

(B)

9%

(C)

10%

(D)

15%

Solution:

Correct Answer is `=>` (C) 10%
Q 2109345218

What is the number of mole(s) of `H_(2) (g)` required to saturate one mole of benzene?
NDA Paper 2 2016
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

`3` moles of `H_(2)(g)` is required to saturate one mole of

benzene.
Correct Answer is `=>` (C) `3`
Q 1760012815

A compound `X_2 O_ 3` contains `31.58%` oxygen by weight. The atomic mass of `X` is
NDA Paper 2 2014
(A)

`34. 66 g mol^(-1)`

(B)

`45.01 g mol^(- 1)`

(C)

`52.00 g mol^(-1)`

(D)

`104.00 g mol^(- 1)`

Solution:

In compound `X_2 O_ 3`,

Percentage of oxygen by weight `= 31.58`

Percentage of `X` by weight `= 68.42`

Let the atomic mass of `X` `= x`

` (2x)/(2x + 48) xx 100 = 68.42`

`:. x = 52`

So, atomic mass of `'X'` is `52 g mol^(- 1)`
Correct Answer is `=>` (C) `52.00 g mol^(-1)`
Q 1731256122

Consider the following reaction `xAs_2 S_ 3 + y O_2 -> zAs_2 O_3 + wSO_2` What is `y` (the coefficient for `O_2` ) when this equation is balanced using whole number coefficients?
NDA Paper 2 2014
(A)

5

(B)

7

(C)

9

(D)

11

Solution:

Balanced chemical equation is

`2 As_2S_ 3 + 9 O_ 2 -> 2 As_2 O_ 3 + 6SO_2`. So, `y` is `9`.
Correct Answer is `=>` (C) 9
Q 1763112045

Note the following balanced chemical equation `2CO + O_2 ⇋ 2CO_2` Which one of the following statements is significant in relation to the above chemical equation?
NDA Paper 2 2014
(A)

One can add to a vessel only `2` mol of `CO` for each mol of `O_2` is added

(B)

No matter how much of these two reagents are added to a vessel, `1` mol of `O_2` is consumed

(C)

When they react, `CO` reacts with `O_2` in a `2 : 1` mol ratio

(D)

When `2` mol of `CO` and `1` mol of `O_2` are placed in a vessel, they will react to give `1` mol of `CO_2`

Solution:

Correct Answer is `=>` (C) When they react, `CO` reacts with `O_2` in a `2 : 1` mol ratio
Q 1781367227

How many grams of `MgCO_3` contain `24.00 g` of oxygen?
(The molar mass of `MgCO_ 3` is `84.30 g mol^(-1) )`

NDA Paper 2 2014
(A)

`42.15 g`

(B)

`84.30 g`

(C)

`126.00 g`

(D)

`154.00 g`

Solution:

In `MgCO_3` , there are `3` oxygen atoms in each

mole.

Atomic mass of oxygen `16 g//rnol`. So there are `(16 xx 3) = 48`

grams of oxygen in one mole of `MgCO_3`

Molar mass of `MgCO_3 = 84.30 g//mol^(-1)`

` (48 g text (of) O)/( 84.3 MgCO_3) = ( 24 g O)/( xg Mg CO_3) `

` x = 42.15 g`
Correct Answer is `=>` (A) `42.15 g`
Q 1711567420

If the reaction of `1.0` mol `NH_ 3 (g)` and `1.0` mol `O_2 (g)` `4NH_3 (g) + 5O_2 (g) -> 4 NO (g) +6 H_2 O (l)` is carried to completion, then
NDA Paper 2 2014
(A)

all the `O_2 (g)` is consumed

(B)

`4.0` mol `NO (g)` is produced

(C)

`1.5` mol `H_2O (l)` is produced

(D)

all the `NH_ 3 (g)` is consumed

Solution:

`4NH_3 + 5O_2 -> 4NO + 6H_2O`

4 mol 5 mol

` 1` mol `NH_3` = `5/4 = 1.25 mol` `O_2`

Since, for `1 mol` of `NH_3 , 1.25 mol` of `O_2` are required,

therefore, `O_ 2` is the limiting factor (here `O_ 2` is only 1 mol).

Hence, all the `O_ 2` will be consumed in reacting with 1 mol

of `NH_3` .
Correct Answer is `=>` (A) all the `O_2 (g)` is consumed
Q 2387356287

Which one among the following equations is correctly balanced?

NDA Paper 2 2012
(A)

`NaOH + Al + H_2O -> 2H_2 + NaAlO_2`

(B)

`2NaOH + 2Al + 2H_2O -> 3H_2 + 2NaAlO_2`

(C)

`2NaOH + 2Al + 3H_2O -> 4H_2 + 2NaAlO_2`

(D)

`2NaOH + 2Al + H_2O -> H_2 + 2NaAlO_2`

Solution:

Among the given equations, the correctly balance equation is `2NaOH + 2Al + 2H_2O -> 3H_2 + 2NaAlO_2`
Correct Answer is `=>` (B) `2NaOH + 2Al + 2H_2O -> 3H_2 + 2NaAlO_2`
Q 2347456383

Which one among the following is the equivalent weight of sulphuric acid? (Atomic weight : `H = 1, S = 32, O = 16`)
NDA Paper 2 2012
(A)

98

(B)

60

(C)

100

(D)

49

Solution:

Equivalent weight of sulphuric acid is 98. Sulphuric acid is a highly corrosive strong mineral acid with the molecular formula `H_2SO_4` . It is a colorless to slightly yellow viscous liquid which is soluble in water at all concentrations. The historical name of this acid is oil of vitriol.
Correct Answer is `=>` (A) 98
Q 2367167085

Which one among the following is the most appropriate statement with respect to the atomic weight of an element?
NDA Paper 2 2012
(A)

The atomic weight of an element is the sum total of the number of protons and neutrons present in the atom of the element

(B)

Unlike mass number, the atomic weight of an element can be a fraction

(C)

The atomic weight of an element is a whole number

(D)

The atomic weight of all the atoms in an element is the same

Solution:

The atomic weight of an element is the sum of number of protons, neutrons and electrons present in the atom of the element. Unlike mass number, it can be a fraction because an element can found in its different isotopes and atomic weight is average of their atomic weight.
Correct Answer is `=>` (B) Unlike mass number, the atomic weight of an element can be a fraction
Q 2347491383

When aqueous solutions of two salts are mixed, the third salt formed may appear as a solid precipitate or a clear solution depending upon the solubility of its ions. It is observed that all salts of `Na, K, NH _4` are soluble. All nitrates and bicarbonates are also soluble. All halides (chlorides, bromides, iodides) are soluble except those of `Ag, Hg (I)` and `Pb`. All sulphates are soluble except those of `Ag, Ca, Ba` and `Pb`. Which one among the following combinations of solutions will produce a solid precipitate'?
NDA Paper 2 2010

(This question may have multiple correct answers)

(A) Sodium sulphate and barium chloride
(B) Magnesium sulphate and barium bicarbonate
(C) Lithium iodide and barium chloride
(D) Ammonium sulphate and potassium bromide
Solution:

According to the given data

`underset(text(Sodium suphate)) (Na_2SO_4)+ underset(text(Barium chloride))(BaCl_2)-> underset(text(Soluble))(2NaCl)+underset(text(White ppt))(BaSO_4) downarrow`

`underset(text(Magnesium sulphate))(MgSO_4)+underset(text(Barium bicarbonate))(Ba(HCO_3)_2)-> underset(text(Soluble))(Mg(HCO_3)_2)+underset(text(White ppt))(BaSO_4) downarrow`

Both the above given combinations of solutions will produce a solid white precipitate of `BaSO_4` Hence, options (1) and (2) both are correct answer.
Correct Answer is `=>` (A)
Q 2337834782

Consider the following equation for the formation of ammonia from nitrogen and hydrogen

`N_2 + 3H_2 ⇋ 2NH_3`

How many hydrogen molecules are required to react with `100` molecules of nitrogen?


NDA Paper 2 2009
(A)

`100`

(B)

`200`

(C)

`300`

(D)

`400`

Solution:

`underset(1 mo l)(N_2)+underset(3 mo l)(3H_2) ⇋ underset(2 mo l)(2NH_3)`

` 1 xx N_0` molecules of `N_2` reacts with `= 3 xx N_0` molecules of `H_2`

`:. 100` molecules of nitrogen will react `= (3 xx N_0 xx 100)/(1 xx N_0)`

`=300` molecules of `H_2`
Correct Answer is `=>` (C) `300`
Q 2367156985

What is the number of water molecules present in a tiny drop of water (volume `0.018` mL) at room temperature'?
NDA Paper 2 2008
(A)

`4.84 xx 10^17`

(B)

`4.184 xx 10^16`

(C)

`6.023 xx 10^19`

(D)

`6.023 xx 10^23`

Solution:

Given, volume of one drop

`=0.018 ` mL`=0.018 xx 10^(-3) L`

`0.018 xx 10^(-3)` L of water at STP

`=(0.018 xx 10^(-3))/(22.4)` mol `=8.035 xx 10^(-7)` mol

Number of molecules in 1 mol water `= 6.023 xx10^23`

`:.` Number of molecules in `8.035 xx 10^(-7)` moles of water

`= 6.023 xx 10^23` `xx 8.035 xx 10^(-7)`

`= 4.839 xx 10^17 = 4.84 xx 10^17`
Correct Answer is `=>` (A) `4.84 xx 10^17`
Q 2317167989

The molecular weight and equivalent weight of which one of the following is the same?
NDA Paper 2 2008
(A)

`H_2SO_4`

(B)

`KMnO_4`

(C)

`H_2C_2O_4`

(D)

`NaOH`

Solution:

Equivalent weight `= (text(Molecular weight))/(text(Acidity(Number of replaceble-OH group))`

For `NaOH`, acidity `= 1`

`:.` Equivalent weight = Molecular weight
Correct Answer is `=>` (D) `NaOH`
Q 2377580486

What is the weight of one atom of hydrogen in gram?
NDA Paper 2 2007
(A)

`6.023xx10^(-23)`

(B)

`1.66 xx 10^(-24)`

(C)

`6.62 xx 10^(-24)`

(D)

none of these

Solution:

Weight of `6.02 xx 10^23` atoms of hydrogen `= 1 g`

`:.` Weight of one atom of hydrogen `=1/(6.02 xx 10^23)`

`=1.66 xx 10^(-24)`
Correct Answer is `=>` (B) `1.66 xx 10^(-24)`
Q 2307580488

Which one of the following elements shows variable equivalent mass'?
NDA Paper 2 2007
(A)

Zinc

(B)

Silver

(C)

Calcium

(D)

Iron

Solution:

Iron (Fe) shows the variable valency hence it shows the variable equivalent mass.
Correct Answer is `=>` (D) Iron
Q 2337880782

The atomic weights are expressed in terms of atomic mass unit. Which one of the following is used as a standard'?
NDA Paper 2 2007
(A)

`text()^1H_1`

(B)

`text()^12 C_6`

(C)

`text()^16 C_8`

(D)

`text()^35 C_17`

Solution:

`text()_6^12 C` is used as a standard in the expression of atomic weights in terms of amu.
Correct Answer is `=>` (B) `text()^12 C_6`
 
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