1
MCQ (Single Correct Answer)

NEET 2013

6.02 $$ \times $$ 1020 molecules of urea present in 100 mL of its solution. The concentration of solution is
A
0.001 M
B
0.1 M
C
0.02 M
D
0.01 M

Explanation

Moles of urea = $${{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}} = 0.001$$

Concentration of solution = $${{0.001} \over {100}} \times 1000$$ = 0.01 M
2
MCQ (Single Correct Answer)

AIPMT 2011 Mains

Which has the maximum number of molecules among the following ?
A
44 g CO2
B
48 g O3
C
8 g H2
D
64 g SO2

Explanation

8 g H2 has 4 moles while the others has 1 mole each.
3
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na+ and carbonate ions, CO$${}_3^{2 - }$$ are respectively (Molar mass of Na2CO3 = 106 g mol$$-$$1)
A
0.955 M and 1.910 M
B
1.910 M and 0.955 M
C
1.90 M and 1.910 M
D
0.477 M and 0.477 M

Explanation

Given in the question that molar mass of Na2CO3 = 106 g

$$ \therefore $$ Molarity of solution = $${{25.3 \times 1000} \over {106 \times 250}}$$

= 0.9547 M = 0.955 M

Na2CO3 $$ \to $$ 2Na+ + $$CO_3^{2-}$$

[Na+] = 2[Na2CO3] = 2 $$ \times $$ 0.955 = 1.910 M

[$$CO_3^{2-}$$] = [Na2CO3] = 0.955 M
4
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

The number of atoms in 0.1 mol of a triatomic gas is (NA = 6.02 $$ \times $$ 1023 mol$$-$$1)
A
6.026 $$ \times $$ 1022
B
1.806 $$ \times $$ 1023
C
3.600 $$ \times $$ 1023
D
1.800 $$ \times $$ 1022

Explanation

No. of atoms

= NA $$ \times $$ No. of moles $$ \times $$ 3

= 6.023 $$ \times $$ 1023 $$ \times $$ 0.1 $$ \times $$ 3

= 1.806 $$ \times $$ 1023

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