JEE Advanced (Subjective Type Questions): Stoichiometry | Chemistry for JEE Main & Advanced PDF Download

Ex.1 Calculate the following for 49 gm of H₂SO₄

(a) moles (b) Molecules (c) Total H atoms (d) Total O atoms (e) Total electrons

 Sol. Molecular wt of H₂SO₄ = 98

(a) moles =  = 

(b) Since 1 mole = 6.023 × 10²³ molecules.

 = 6.023 × 10²³ × molecules = 3.011 × 10²³ molecules

(c) 1 molecule of H₂SO₄ Contains 2 H atom

3.011 × 10²³ of H₂SO₄ contain 2 × 3.011 × 10²³ atoms = 6.023 × 10²³ atoms

(d) 1 molecules of H₂SO₄ contains 4 O atoms

3.011 × 10²³ molecular of H₂SO₄ contains = 4 × 3.011 × 10²³ = 12.044 × 10²³

(e) 1 molecule of H₂SO₄ contains 2H atoms 1 S atom 4 O atom

this means 1 molecule of H₂SO₄ Contains (2 16 4 × 8) e^_

So 3.011 × 10²³ molecules have 3.011 × 10²³ × 50 electrons = 1.5055 × 10²⁵ e^_

Ex.2 Calculate the total ions & charge present in 4.2 gm of N^_3

 Sol. mole =  =  = 0.3

total no of ions = 0.3 × N_A ions

total charge = 0.3 N_A × 3 × 1.6 × 10^_19

= 0.3 × 6.023 × 10²³ × 3 × 1.6 × 10^_19 , = 8.67 × 10⁴ C Ans.

Ex.3 Find the total number of iron atom present in 224 amu iron.

 Sol. Since 56 amu = 1 atom

therefore 224 amu =  × 224 = 4 atom Ans.

Ex.4 A compound containing Ca, C, N and S was subjected to quantitative analysis and formula mass determination. A 0.25 g of this compound was mixed with Na₂CO₃ to convert all Ca into 0.16 g CaCO₃. A 0.115 gm sample of compound was carried through a series of reactions until all its S was changed into SO₄^2_ and precipitated as 0.344 g of BaSO₄. A 0.712 g sample was processed to liberated all of its N as NH₃ and 0.155 g NH₃ was obtained. The formula mass was found to be 156. Determine the empirical and molecular formula of the compound.

 Sol. Moles of CaCO₃ =  = Moles of Ca

Wt of Ca =  × 40

Mass % of Ca = 

Similarly Mass % of S = 

Similarly Mass % of N = = 17.9

Þ Mass % of C = 15.48

Now :

Elements Ca S N C

Mass % 25.6 41 17.9 15.48

Mol ratio 0.64 1.28 1.28 1.29

Simple ratio 1 2 2 2

Empirical formula = CaC₂N₂S₂,

Molecular formula wt = 156 , n × 156 = 156 Þ n = 1

Hence, molecular formula = CaC₂N₂S₂

Ex.5 A polystyrne having formula Br₃C₆H₃(C₃ H₈)_n found to contain 10.46% of bromine by weight. Find the value of n. (At. wt. Br = 80)

 Sol. Let the wt of compound is 100 gm & molecular wt is M

Then moles of compound =

Moles of Br = × 3

wt of Br = × 3 × 80 = 10.46

M = 2294.45 = 240 75 44 n , Hence n = 45 Ans.

Ex.6 A sample of clay was partially dried and then analysed to 50% silica and 7% water. The original clay contained 12% water. Find the percentage of silica in the original sample.

 Sol. In the partially dried clay the total percentage of silica water = 57%. The rest of 43% must be some impurity. Therefore the ratio of wts. of silica to impurity = . This would be true in the original sample of silica.

The total percentage of silica impurity in the original sample is 88. If x is the percentage of silica, ; x = 47.3% Ans.

Ex.7 A mixture of CuSO₄.5H₂O and MgSO₄. 7H₂O was heated until all the water was driven-off. if 5.0 g of mixture gave 3 g of anhydrous salts, what was the percentage by mass of CuSO₄.5H₂O in the original mixture ?

 Sol. Let the mixture contain x g CuSO₄.5H₂O

Þ = 3 Þ x = 3.56

Þ Mass percentage of CuSO₄. 5H₂O = = 71.25 % Ans.

Ex .8 367.5 gm KClO₃ (M = 122.5) when heated, How many litre of oxygen gas is proudced at S.T.P.

 Sol. KClO₃ ® KCl O₂

Applying POAC on O, moles of O in KClO₃ = moles of O in O₂

3 × moles of KClO₃ = 2 × moles of O₂

3 × = 2 × n, n = ×

Volume of O₂ gas at S.T.P = moles × 22.4

=  = 9 × 11.2 = 100.8 lit Ans.

Ex.9 0.532 g of the chloroplatinate of a diacid base on ignition left 0.195 g of residue of Pt. Calculate molecular weight of the base (Pt = 195)

 Sol. Suppose the diacid base is B.

B H₂PtCl₆® BH₂PtCl₆ ® Pt

diacid acid chloroplatinate

base 0.532 g 0.195 g

Since Pt atoms are conserved, applying POAC for Pt atoms,

moles of Pt atoms in BH₂PtCl₆ = moles of Pt atoms in the product

1 × moles of BH₂PtCl₆ = moles of Pt in the product

mol. wt. of BH₂PtCl₆ = 532

From the formula BH₂PtCl₆, we get

mol. wt. of B = mol. wt. of BH₂PtCl₆ _ mol. wt. of H₂PtCl₆

= 532 _ 410 = 122. Ans.

Ex.10 10 mL of a gaseous organic compound containing. C, H and O only was mixed with 100 mL of oxygen and exploded under conditions which allowed the water formed to condense. The volume of the gas after explosion was 90 mL. On treatment with potash solution, a further contraction of 20 mL in volume was observed. Given that the vapour density of the compound is 23, deduce the molecular formula. All volume measurements were carried out under the same conditions.

Sol. C_xH_yO_z    O₂ ® xCO₂   H₂O

10 ml

after explosion volume of gas = 90 ml

90 = volume of CO₂ gas volume of unreacted O₂

on treatment with KOH solution volume reduces by 20 ml. This means the volume of CO₂ = 20 ml

the volume of unreacted O₂ = 70 ml

volume of reacted O₂ = 30 ml

V.D of compoud = 23

molecular wt 12x y 16z = 46 ...(1)

from equation we can write

, x   _  = 3

4x y _ 2z = 12 ...(2)

& 10x = 20 Þ x = 2

from eq. (1) & (2) ; z = 1 & y = 6; Hence C₂H₆O Ans.