$$ \therefore $$ It would change the mass of one mole of carbon.
2
AIPMT 2015
MCQ (Single Correct Answer)
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
$$ \Rightarrow $$ w = (nAgCl) $$ \times $$ Molecular mass
= (0.049) $$ \times $$ (107.8 + 35.5) = 7.02 g
3
AIPMT 2014
MCQ (Single Correct Answer)
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
4
AIPMT 2014
MCQ (Single Correct Answer)
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 = $${{Mass} \over {Mol.\,\,mass}}$$