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**Ex.1** **Calculate the following for 49 gm of H _{2}SO_{4}**

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

**Sol. **Molecular wt of H_{2}SO_{4} = 98

(a) moles = =

(b) Since 1 mole = 6.023 × 10^{23} molecules.

= 6.023 × 10^{23} × molecules = 3.011 × 10^{23} molecules

(c) 1 molecule of H_{2}SO_{4} Contains 2 H atom

3.011 × 10^{23} of H_{2}SO_{4} contain 2 × 3.011 × 10^{23} atoms = 6.023 × 10^{23} atoms

(d) 1 molecules of H_{2}SO_{4} contains 4 O atoms

3.011 × 10^{23} molecular of H_{2}SO_{4} contains = 4 × 3.011 × 10^{23} = 12.044 × 10^{23}

(e) 1 molecule of H_{2}SO_{4} contains 2H atoms 1 S atom 4 O atom

this means 1 molecule of H_{2}SO_{4} Contains (2 16 4 × 8) e^{_}

So 3.011 × 10^{23} molecules have 3.011 × 10^{23} × 50 electrons = 1.5055 × 10^{25} 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 × 10^{23} × 3 × 1.6 × 10^{_19} , = 8.67 × 10^{4} 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 Na_{2}CO_{3} to convert all Ca into 0.16 g CaCO_{3}. A 0.115 gm sample of compound was carried through a series of reactions until all its S was changed into SO_{4}^{2_} and precipitated as 0.344 g of BaSO_{4}. A 0.712 g sample was processed to liberated all of its N as NH_{3} and 0.155 g NH_{3} was obtained. The formula mass was found to be 156. Determine the empirical and molecular formula of the compound.

**Sol. **Moles of CaCO_{3} = = 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 = CaC_{2}N_{2}S_{2},

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

Hence, molecular formula = CaC_{2}N_{2}S_{2}

**Ex.5 A polystyrne having formula Br _{3}C_{6}H_{3}(C_{3} H_{8})_{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 CuSO _{4}.5H_{2}O and MgSO_{4}. 7H_{2}O 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 CuSO_{4}.5H_{2}O in the original mixture ?**

**Sol. **Let the mixture contain x g CuSO_{4}.5H_{2}O

Þ = 3 Þ x = 3.56

Þ Mass percentage of CuSO_{4}. 5H_{2}O = = 71.25 % **Ans.**

**Ex .8** **367.5 gm KClO _{3} (M = 122.5) when heated, How many litre of oxygen gas is proudced at S.T.P.**

**Sol.** KClO_{3} ® KCl O_{2}

Applying POAC on O, moles of O in KClO_{3} = moles of O in O_{2}

3 × moles of KClO_{3} = 2 × moles of O_{2}

3 × = 2 × n, n = ×

Volume of O_{2} 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 H_{2}PtCl_{6}® BH_{2}PtCl_{6} ® 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 BH_{2}PtCl_{6} = moles of Pt atoms in the product

1 × moles of BH_{2}PtCl_{6} = moles of Pt in the product

mol. wt. of BH_{2}PtCl_{6} = 532

From the formula BH_{2}PtCl_{6}, we get

mol. wt. of B = mol. wt. of BH_{2}PtCl_{6} _ mol. wt. of H_{2}PtCl_{6}

= 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_{x}H_{y}O_{z} O_{2} ® xCO_{2} H_{2}O

10 ml

after explosion volume of gas = 90 ml

90 = volume of CO_{2} gas volume of unreacted O_{2}

on treatment with KOH solution volume reduces by 20 ml. This means the volume of CO_{2} = 20 ml

the volume of unreacted O_{2} = 70 ml

volume of reacted O_{2} = 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 C_{2}H_{6}O **Ans.**

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