1
MCQ (Single Correct Answer)

AIPMT 2008

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
A
CHO
B
CH4O
C
CH3O
D
CH2O

Explanation

Element % Atomic mass mole ratio simple ratio
C 38.71 12 $${{38.71} \over {12}} = 3.22$$ $${{3.22} \over {3.22}} = 1$$
H 9.67 1 $${{9.67} \over 1} = 9.67$$ $${{9.67} \over {3.22}} = 1$$
O 51.62 16 $${{51.62} \over {16}} = 3.22$$ $${{3.22} \over {3.22}} = 1$$

Hence empirical formula of the compound would be CH3O
2
MCQ (Single Correct Answer)

AIPMT 2008

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g HCl ?
A
0.011
B
0.029
C
0.044
D
0.333

Explanation

PbO + 2HCl $$\to$$ PbCl2 + H2O
n mol 2n mol n mol
$$6.5 \over 224$$ mol $$3.2 \over 36.5$$ mol
0.029 mol 0.087 mol

Formation of moles of lead(II) chloride depends upon the no. of moles of PbO which acts as a limiting reagent here. So no of moles of PbCl2 formed will be equal to the no of PbO (i.e 0.029)
3
MCQ (Single Correct Answer)

AIPMT 2007

An element, X has the following isotopic composition:
200X : 90%  199X : 8.0%  202X : 2.0%

The weighted average atomic mass of the naturally occuring element X is closest to
A
201 amu
B
202 amu
C
199 amu
D
200 amu

Explanation

Average isotope mass of X

= $${{200 \times 90 + 199 \times 8 + 202 \times 2} \over {90 + 8 + 2}}$$

= $${{18000 + 1592 + 404} \over {100}}$$

= 199.96 a.m.u $$ \approx $$ 200 a.m.u
4
MCQ (Single Correct Answer)

AIPMT 2004

The maximum number of molecules is present in
A
15 L of H2 gas at STP
B
5 L of N2 gas at STP
C
0.5 g of H2 gas
D
10 g of O2 gas

Explanation

At STP, 22.4 L H2

= 6.023 $$ \times $$ 1023 molecules

15 L H2 = $${{6.023 \times {{10}^{23}} \times 15} \over {22.4}}$$ = $$4.033 \times {10^{23}}$$

5 L N2 = $${{6.023 \times {{10}^{23}} \times 5} \over {22.4}}$$ = 1.344 $$ \times $$ 1023

2 g of H2 = 6.023 $$ \times $$ 1023

0.5 g H2 = $${{6.023 \times {{10}^{23}} \times 0.5} \over {2}}$$ = 1.505 $$ \times $$ 1023

32 g O2 = 6.023 $$ \times $$ 1023

10 g O2 = $${{6.023 \times {{10}^{23}} \times 10} \over {32}}$$ = 1.882 $$ \times $$ 1023

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