Class 11th Chemistry Solution Chapter 1

NCERT Solutions for Class 11th Chemistry

Class 11 Chemistry SOLUTION

Chapter 1 Some basic Concepts Of Chemistry

Question 1. Calculate the molecular mass of the following:
(i) H20(ii) C02(iii) CH4

Answer:  (i) Molecular mass of H2O = 2(1.008 amu) + 16.00 amu=18.016 amu
(ii) Molecular mass of CO2= 12.01 amu + 2 x 16.00 amu = 44.01 amu
(iii) Molecular mass of CH4= 12.01 amu + 4 (1.008 amu) = 16.042 amu

Question 2. Calculate the mass percent of different elements present in sodium sulphate (Na2 SO4).
Answer:

1.3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Answer

% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]

Atomic mass of iron = 55.85 amu

Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88

Simplest molar ratio = 1.25/1.25 : 1.88/1.25 

 ⇒ 1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe₂O₃.

1.4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer

The balanced reaction of combustion of carbon in dioxygens is:
C(s)     +    O₂(g)     →    CO₂  (g)
1mole     1mole(32g)     1mole(44g)

(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide  produced by burning 1 mole of carbon.
(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again oxgen acts as a limiting reagent as only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

1.5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245g mol^-1.

Answer

0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.

∴No. of moles of sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875

Molar mass of sodium acetate = 82.0245g mol^-1

∴Mass of sodium acetate acquired =  0.1875×82.0245 g = 15.380g

1.6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^-1 and the mass per cent of nitric acid in it being 69%.

Answer

Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO₃) = 1+14+48 = 63g mol^-1
Number of moles in 69 g of HNO₃ = 69/63 moles = 1.095 moles
Volume of 100g  nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L
∴ Conc. of HNO₃ in moles per litre = 1.095/0.07092 = 15.44 M

1.7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄ )?

Answer

1 mole of CuSO₄ contains 1 mole of copper.
Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
∴ copper can be obtained from 100 g of copper sulphate = (63.5/159.5)×100 = 39.81g

8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.8 g mol-1(Atomic mass: Fe = 55.85, O = 16.00 amu)Calculation of Empirical Formula. See Q3.

Answer: Empirical formula mass of Fe203 = 2 x 55.85 + 3 x 16.00 = 159.7 g mol–1

Hence, molecular formula is same as empirical formula, viz.,Fe203.

Question 9.Calculate the atomic mass (average) of chlorine using the following data:

Answer:

Question 10.In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms (ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane
Answer: (i) 1 mole of C2H6 contains 2 moles of carbon atoms
.•. 3 moles of C2H6 will C-atoms = 6 moles
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms
.•. 3 moles of C2H6 will contain H-atoms = 18 moles

Question 11. What is the concentration of sugar (C12H22O11) in mol L -1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Answer:

Question 12. If the density of methanol is 0.793 kg L -1, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer:

Question 13. Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal, is as shown below:1 Pa = 1 Nm-2.If mass of air at sea level is 1034 g cm-2,calculate the pressure in pascal.

Answer: Pressure is the force (i.e., weight) acting per unit area But weight = mg

Question 14. What is the S.I. unit of mass?
Answer:  S.I. unit of mass is kilogram (kg).

15. Match the following prefixes with their multiples:

     Prefixes                        Multiples

(i)  micro                              10⁶

(ii) deca                                10⁹

(iii) mega                              10^-6

(iv) giga                                10^-15

(v) femto                               10

Answer

 Prefixes                        Multiples

(i)  micro                              10^-6

(ii) deca                                10

(iii) mega                              10⁶

(iv) giga                                10⁹

(v) femto                               10^-15

1.16. What do you mean by significant figures ?

Answer

Significant figures are meaningful digits which are known with certainty including the last digit whose value is uncertain.

For example, 

In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the uncertainty would be ±1 in the last digit. Hence last uncertain digit is also included in Significant figures.

1.17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.

Answer

(i) 15 ppm means 5 parts in million(10⁶⁾ parts.
∴ % by mass = 15/10⁶ × 100 = 15×10^-4 = 1.5×10^-3 %
(ii) Molar mass of chloroform(CHCl₃) = 12+1+(3×35.5) = 118.5 g mol^-1
100g of the sample contain chloroform = 1.5×10^-3g
∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5×10^-2 g
= 1.5×10^-2/118.65 mole = 1.266×10^-4 mole
∴ Molality = 1.266×10^-4 m.

1.18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Answer

(i) 0.0048 = 4.8×10^-3
(ii) 234, 000 = 2.34×10⁵
(iii) 8008 = 8.008×10³
(iv) 500.0 = 5.000×10²
(v) 6.0012 = 6.0012×10⁰

1.19. How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034

Answer

(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5

1.20. Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808

Answer

(i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810

Question 20. Round up the following upto three significant figures:
(i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808
Answer: (i) 34.2 (ii) 10.4 (iii) 0.0460 (iv) 2810

Question 21. The following data were obtained when dinitrogen and dioxygen react together to form compounds:

Answer: (a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32,64, 32 and 80 g in the given four oxides. These’are in the ratio 1 : 2 : 1 : 5 which is a simple whole number ratio. Hence, the given data obey the law of multiple proportions.

Question 22.

Answer:

Question 23. In the reaction, A + B2——> AB2, identify the limiting reagent, if any, in the following mixtures
(i) 300 atoms of A + 200 molecules ofB
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules ofB
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Answer:  (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
.•. 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be
left unreacted. Hence, B is the limiting reagent while A is the excess reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
.•. 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant.
(iii) No limiting reagent.
(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.
(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent.

1.24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N₂(g) + H₂(g) → 2NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00×10³g dinitrogen reacts with 1.00×10³g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

Answer

1 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g).

∴ 2000 g of N₂ will react with H₂ = 6/28 ×200g = 428.6g. Thus, here N₂ is the limiting reagent while H₂ is in excess.

28g of N₂ produce 34g of NH_3.

∴2000g of N₂ will produce = 34/28×2000g = 2428.57 g of NH_3.

(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1000g - 428.6g = 571.4 g

1.25. How are 0.50 mol Na₂CO₃and 0.50 M Na₂CO₃ different?

Answer

Molar mass of Na₂CO₃ = (2×23)+12.00+(3×16) = 106 g mol^-1

∴0.50 mol Na₂CO₃ means 0.50×106g = 53g

0.50 M Na₂CO₃ means 0.50 mol of Na₂CO₃ i.e. 53g of  Na₂CO₃ are present in 1litre of the solution.

1.26. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer

Dihydrogen gas reacts with dioxygen gas as, 

2H₂(g) + O₂(g) → 2H₂O(g)

Thus, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

1.27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Answer

(i) 1 pm = 10^-12 m
28.7 pm = 28.7×10^-12 m = 2.87×10^-11 m
(ii) 1 pm = 10^-12 m
∴15.15 pm = 15.15×10^-12 m = 1.515 ×10^-11 m
(iii) 1 mg = 10^-3 g
25365 mg = 2.5365×10⁴×10^-3 g
Now,
1 g = 10^-3 kg
2.5365×10 g = 2.5365×10×10^-3 kg
∴25365 mg = 2.5365×10^-2 kg

1.28. Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl₂ (g)

Answer

(i) 1 g Au = 1/197 mol = 1/197×6.022×10²³ atoms
(ii) 1 g Na = 1/23 mol = 1/23×6.022×10²³ atoms
(iii) 1 g Li = 1/7 mol = 1/7×6.022×10²³ atoms
(iv) 1 g Cl₂ = 1/71 mol = 1/71×6.022×10²³ atoms
Thus, 1 g of Li has the largest number of atoms.

1.29. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Answer

Mole fraction of C₂H₅OH = No. of moles of C₂H₅OH/No. of moles of solution
n_C2H5OH = n(C₂H₅OH)/(C₂H₅OH)+n(H₂O) = 0.040 (Given) ... 1
We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute. Therefor, water is approx. 1L.
No. of moles in 1L of water = 1000g/18g mol^-1 = 55.55 moles
Substituting n(H₂O) = 55.55 in equation 1
n(C₂H₅OH)/(C₂H₅OH) + 55.55 = 0.040
⇒ 0.96n(C₂H₅OH) = 55.55 × 0.040
⇒ n(C₂H₅OH) = 2.31 mol
Hence, molarity of the solution = 2.31M

1.30. What will be the mass of one  ¹²C atom in g ?

Answer

1 mol of ¹²C atoms = 6.022×10²³ atoms = 12g
∴ Mass of 1 atom ¹²C = 12/6.022×10²³ g = 1.9927×10^-23 g

Question 31. How many significant figures should be present in the answer of the following?

Answer:  (i) The least precise term has 3 significant figures (i.e., in 0.112). Hence, the answer should have 3 significant figures.
(ii) Leaving the exact number (5), the second term has 4 significant figures. Hence, the answer should have 4 significant figures.
(iii) In the given addition, the least number of decimal places in the term is 4. Hence, the answer should have 4 significant.

Question 32. Use the data given in the following table to calculate the molar mass of naturally occurring argon.

Answer: Molar mass of Ar = 35.96755 x 0.00337 + 37.96272 x 0.00063 + 39.96924 x 0.99600 = 39.948 g mol-1

Question 33. Calculate the number of atoms in each of the following:
(i) 52 moles of He (ii) 52 u of He (iii) 52 g of He
Answer:

Question 34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Answer:

Question 35. Calcium carbonate reacts with aqueous HCl according to the reaction

CaC03 (s) + 2HCl (aq) ———->CaCl2 (aq) +C02(g) +H2O(l).

What mass of CaC03 is required to react completely with 25 mL of 0.75 M HCl?
Answer: Step 1. To calculate mass of HCl in 25 mL of 0.75 m HCl
1000 mL of 0.75 M HCl contain HCl = 0.75 mol = 0.75 x 36.5 g = 24.375 g

Step 2. To calculate mass of CaC03reacting completely with 0.9125 g of HCl
CaC03 (s) + 2HC1 (aq)———->CaCl2(aq) +C02(g) + H2O
2 mol of HCl, i.e., 2 x 36.5 g = 73 g HCl react completely with CaC03 = 1 mol = 100 g

Question 36. Chlorine is prepared in the laboratory by treating manganese dioxide (Mn02) with aqueous hydrochloric acid according to the reaction.
4 HCl (aq) + Mn02 (s) ———–> 2 H2O (l) + MnCl2(aq) +Cl2(g)

How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 u)
Answer: 1 mole of Mn02, i.e., 55 + 32 = 87 g Mn02 react with 4 moles of HCl, i.e., 4 x 36.5 g = 146 g of HCl.

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