1
MCQ (Single Correct Answer)

AIPMT 2015

What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5
A
3.5 g
B
7 g
C
14 g
D
28 g

Explanation

50 ml of 16.9% solution of AgNO3

$$\left( {{{16.9} \over {100}} \times 50} \right)$$ = 8.45 g of AgNO3

nmole = $${{8.45g} \over {(107.8 + 14 + 16 \times 3)g/mol}}$$

= $$\left( {{{8.45g} \over {169.8g/mol}}} \right) = 0.0497\,moles$$

50ml of 5.8% solution of NaCl contain

NaCl = $$\left( {{{5.8} \over {100}} \times 50} \right) = 2.9g$$

nNaCl = $${{2.9g} \over {(23 + 35.5)g/mol}}$$

= 0.0495 moles

AgNO3 + NaCl $$ \Rightarrow $$ AgCl + Na$$ \oplus $$ + Cl$$ \ominus $$
1 mole 1 mole 1 mole
0.049 mole 0.049 mole 0.049 mole of AgCl

n = $${w \over M}$$

$$ \Rightarrow $$ w = (nAgCl) $$ \times $$ Molecular mass

= (0.049) $$ \times $$ (107.8 + 35.5) = 7.02 g
2
MCQ (Single Correct Answer)

AIPMT 2014

1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel, Which reactant is left in excess and how much ? (At. wt. Mg = 24, O = 16)
A
Mg. 0.16 g
B
O2, 0.16 g
C
Mg, 0.44 g
D
O2 , 0.28 g

Explanation

nMg = $$1 \over 24$$ = 0.0416 moles

nO2 = $$0.56 \over 32$$ = 0.0175 moles

Mg + $${1\over2} O_2$$ $$ \Rightarrow $$ MgO
Initial 0.0416 moles 0.0175 moles 0
Final ( 0.0416 - 2 x 0.0175)
= 0.0066 moles
0 2 x 0.0175

$$ \because $$ Mass of Mg left in excess = 0.0066 $$ \times $$ 24 = 0.16 g
3
MCQ (Single Correct Answer)

AIPMT 2014

Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27oC in identical conditions. The ratio of the volumes of gases H2 : O2 : methane would be
A
8 : 16 : 1
B
16 : 8 : 1
C
16 : 1 : 2
D
8 : 1 : 2

Explanation

According to Avogadro's hypothesis ratio of the volumes of gases will be equal to the ratio of their no. of moles.

So, no of moles = $${{Mass} \over {Mol.\,\,mass}}$$

nH2 = $${w \over 2}$$; nO2 = $${w \over 32}$$; nCH4 = $${w \over 16}$$

So, the ratio will be $${w \over 2}$$ : $${w \over 32}$$ : $${w \over 16}$$ or 16 : 1 : 2
4
MCQ (Single Correct Answer)

AIPMT 2014

When 22.4 liters of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P. the moles of HCl(g) formed is equal to
A
1 mol of HCl(g)
B
2 mol of HCl(g)
C
0.5 mol of HCl(g)
D
1.5 mol of HCl(g)

Explanation

1 mole = 22.4 L at S.T.P

nH2 = $${{22.4} \over {22.4}}$$ = 1 mole

nCl2 = $${{11.2} \over {22.4}}$$ = 0.5 mole

H2 + Cl2 $$ \Rightarrow $$ 2HCl(g)
Initial 1 mol 0.5 mol 0
Final ( 1 - 0.5)
= 0.5 mol
(0.5 - 0.5)
= 0 mol
2 x 0.5
1 mol

Here Cl2 is limiting reagent. So 1 mole of HCl(g) is formed.

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN