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(i)CuSO₄ + 2NH₄OH ⟶ Cu(OH)₂ ↓ (NH₄)₂ SO₄

Blue pale blue ppt. colourless is solution

With excess of NH₄OH, ppt dissolves

CU(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Cu(NH₃)₄]SO₄ + 4H₂O

Excess Tetrammine

Copper(II) Sulphate

(ii)ZnSO₄ + 2NH₄OH ⟶ Zn(OH)₂ + (NH₄)₂ SO₄

Colourless white, gelatinous ppt colourless

With excess of NH₄OH, ppt dissolves

Zn(OH)₂ + (NH₄)2SO₄ + 2NH₄OH ⟶ [Zn(NH₃)₄]SO₄ + 4H₂O

(excess) Tetramminezinc(II) Sulphate

(colourless)

• Molarity is the number of moles of the solute present in 1 litre of the solution. Therefore, molarity depends on the volume of the solution.
• Volume of the solution varies with the change in temperature.

Hence,

Molarity of a solution depends upon temperature.

The branch of chemistry which deals with the study of separation, identification, qualitative and quantitative determination of the compositions of different substances, is called analytical chemistry.

1. The accuracy of measurement is of great concern in analytical chemistry. This is because faulty equipment, poor data processing, or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in analytical measurement.

2. When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.

3. Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.

4. Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation, and properly express the quantitative error in the result.

a. 26 mL 

b. 0.0025 m 

c. 1.5 × 10⁵ mg 

d. 2.0 × 10²g

To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent.

The coefficients are then added/subtracted.

(a). (0.23 × 10⁴mL) + (4.22 × 10⁴mL) +(0.904 × 10⁴mL) + (87.1 × 10⁴mL)

= (0.23 + 4.22 + 0.904 + 87.1) × 10⁴mL 

= 92.454 × 10⁴mL 

= 9.2454 × 10⁵ 

= 9.2 × 10⁵mL

(b). 319.5 g – 20460 g – 0.0639 g – 45.642 g – 4.173 g

= – 20190.3789 g 

= – 20190 g

∴ Sum to appropriate number of significant figures = 9.2 × 10⁵mL

Sum to appropriate number of significant figures = – 20190 g

[Note : In addition and subtraction, the final answer is rounded to the minimum number of decimal point of the number taking part in calculation. If there is no decimal point, then the final answer will have no decimal point.]

a. 4 

b. 4 

c. 3 

d. 3

a. 4

b. 3 

c. 2 

d. 5 

e. 3 

f. 2 

g. 3 

h. 4

a. 0.0003498 = 3.498 × 10^-4

b. 235.4678 = 2.354678 × 10²

c. 70000.0 = 7.00000 × 10⁴

d. 1569.00 = 1.56900 × 10³

Zn(OH)₂ is soluble in excess of NaOH.

a. 3.49 × 10^-11 = 0.0000000000349

b. 3.75 × 10^-1 = 0.375

c. 5.16 × 10⁴ = 51,600

d. 43.71 × 10^-4 = 0.004371

e. 0.011 × 10^-3 = 0.000011

f. 14.3 × 10^-2 = 0.143

g. 0.00477 × 10⁵ = 477

h. 5.00858585 = 5.00858585

a. 3,672,199 = 3.672199 × 10⁶

b. 0.000098 = 9.8 × 10^-5

c. 0.00461 = 4.61 × 10^-3

d. 198.75 = 1.9875 × 10²

Option : b. 36 % 

Mg(OH)₂ [Because all others are amphoteric hydroxides]

 Zn²⁺ is white hydroxide which dissolves in an excess of either aqueous sodium hydroxide or ammonium hydroxide.

Option : c. CH₃

Option : d. A₂B₃

Option : c.14.0 g

Option : c. 24

Option : b. 27

(i) Fe(OH)₂ and Pb(OH)₂

(ii) Cu(OH)₂ and Zn(OH)₂

(i) Ca(OH)₂

(ii) Fe(OH)₂ and Cu(OH)₂

(iii) Zn(OH)₂ and Pb(OH)₂

Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

(a) PbO

(b) ZnO

(c) K₂ZnO₂

Zinc, aluminium and lead; acids and alkalis 

When freshly precipitated aluminum hydroxide reacts with caustic soda solution, whitesalt of sodium meta aluminate is obtained.

Al(OH)₃ + NaOH → NaAlO₂ + 2H₂O

Sodium meta aluminate

(a) Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

(b) Zn + 2NaOH → Na₂ZnO₂ + H₂

1. ZnO + 2NaOH → Na₂ ZnO₂ + H₂O

2. Zn + 2NaOH → Na₂ ZnO₂ + H₂↑

Given : 

i. Atomic mass : 

Fe = 56; 

S = 32; 

O = 16 

ii. Mass of crystal = 4.54 kg 

To find :

i. Mass percentage of Fe, S, H and O 

ii. Mass of iron and water of crystallisation in 4.54 kg of crystal

Formula :

Percentage (by weight) = \(\frac{Mass\,of\,the\,element\,in\,1\,mole\,of\,compound}{Molar\,mass\,of\,the\,compound}\) \(\times 100\)

i. Molar mass of FeSO₄.7H₂O 

= 1 × (56) + 1 × (32) + 14 × (1) + 11 × (16) 

= 56 + 32 + 14+ 176 

= 278 g mol^-1

Percentage of Fe = \(\frac{56}{278}\) \(\times 100\)

= 20.14%

Percentage of S = \(\frac{32}{278}\)\(\times 100\)

= 11.51%

Percentage of H = \(\frac{14}{278}\)\(\times 100\)

= 5.04%

Percentage of O = \(\frac{176}{278}\)\(\times 100\)

= 63.31%

ii. 278 kg green vitriol 

= 56 kg iron 

∴ 4.54 kg green vitriol = x 

∴ x = \(\frac{56\times 4.54}{278}\)

Mass of 7H₂O in 278 kg green vitriol = 7 × 18

= 126 kg

∴ 4.54 kg green vitriol = y

∴ y = \(\frac{126\times 4.54}{278}\)

∴ i. Mass percentage of Fe, S, H and O in FeSO₄ .7H₂O are 20.14%, 11.51%, 5.04% and 63.31% respectively.

ii. Mass of iron in 4.54 kg green vitriol = 0.915 kg

Mass of water of crystallisation in 4.54 kg 

green vitriol = 2.058 kg

Accuracy : 

1. Accuracy refers to nearness of the measured value to the true value. 

2. Accuracy represents the correctness of the measurement. 

3. Accuracy is expressed in terms of absolute error and relative error. 

4. Accuracy takes into account the true or accepted value. 

5. Accuracy can be determined by a single measurement. 

6. High accuracy implies smaller error.

Precision : 

1. Precision refers to closeness of multiple readings of the same quantity. 

2. Precision represents the agreement between two or more measured values. 

3. Precision is expressed in terms of absolute deviation and relative deviation.

4. Precision does not take into account the true or accepted value. 

5. Several measurements are required to determine precision. 

6. High precision implies reproducibility of the readings.

(i) FeCI₃ + 3NaOH ⟶ Fe(OH)₃ ↓ 3 NaCI

Yellow reddish brown, ppt colourless in solution

In excess of alkali, the reddish brown ppt, of Fe(OH)₃ remains insoluble

(ii) ZnSO₄ + 2NaOH ⟶ Zn(OH)₂ ↓ + NaSO₄

Colourless white gelatinous ppt. colourless

In excess of alkali, white gelatinous ppt. of Zn(OH)₂ becomes soluble

Zn(OH)₂ + 2NaOH (Excess) ⟶ Na₂ZnO₂ + 2H₂O

Sodium zincate (colourless)

(iii) Pb(NO₃)₂ + 2NaOH ⟶ Pb(OH)₂ ↓ + 2NaNO₃

White ppt (colourless)

In excess of alkali, white precipitate of Pb(OH)₂ becom essoluble:

Pb(OH)₂ + 2NaOH(excess) ⟶ Na₂PbO₂ + 2H₂O

Sodium plumbate

{colourless}

CuSO₄ + 2NaOH ⟶ Cu(OH)₂↓ + 2NaSO₄

Blue colourless pale blue ppt. { colourless}

In excess of alkali, pale blue precipitate of Cu(OH)₂ is insoluble

Reagent bottles A and B can identified by using calcium salts such as Ca(NO₃)₂.

On adding NaOH to Ca (NO₃)₂, Ca (OH) ₂ is precipitated as white precipitate which is sparingly soluble in excess of NaOH.

Ca(NO₃)₂ + 2NaOH⟶ Ca(OH)₂ + 2NaNO₃

Whereas, on addition of NH₄OH to calcium salts, no precipitation of Ca(OH)₂ occurs even with addition of excess of NH₄OH because the concentration of OH^-ions from the ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.

So the reagent bottle which gives white precipitate is NaOH and the other is NH₄OH.

(i) Analysis: The determination of chemical components in a given sample is called analysis.

(ii) Qualitative analysis: The analysis which involves the identification of the unknown substances in a given sample is called qualitative analysis.

(iii) Reagent: A reagent is a substance that reacts with another substance.

(iv) Precipitation: It is the process of formation of an insoluble solid when solutions are mixed. The solid thus formed is called precipitate.

Option : b. 3

Option : c. Colour

Option : a. kg

i. In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI units, abbreviated from its French name.

ii. The SI unit of mass is kilogram (kg).

Limiting reagent :

• The reactant which gets consumed and limits the amount of product formed is called the limiting reagent.
• When a chemist carries out a reaction, the reactants are not usually present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation.
• This is because the goal of a reaction is to produce the maximum quantity of a useful compound from the starting materials. Frequently, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted into the desired product.
• The reactant which is present in lesser amount gets consumed after some time and subsequently, no further reaction takes place, whatever be the amount left of the other reactant present.

Hence,

Limiting reagent is the reactant that gets consumed entirely and limits the reaction.

a. (3.26 × 10⁴) (1.54 × 10⁶) = 5.0204 × 10⁴⁺⁶ 

= 5.02 × 10¹⁰

b. (8.39 × 10⁷) (4.53 × 10⁹) = 38.0067 × 10⁷⁺⁹

= 38.0067 × 10¹⁶ 

= 3.80 x 10¹⁷

c. \(\frac{8.94\times 10^6}{4.35\times 10^4}\) = 2.055 x 10^6-4

= 2.06 x 10²

d. \(\frac{(9.28\times 10^9)\times (9.9\times 10^{-7})}{(511)\times (2.98\times 10^{-6})}\) = 0.06033 x 10^9-7-(-6)

= 0.06033 x 10⁸

= 6.03 x 10⁶

To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. 

The coefficients are then added/subtracted.

a. 3.971 × 10⁷ + 1.98 × 10⁴ 

= 3.971 × 10⁷ + 0.00198 × 10⁷ 

= (3.971 + 0.00198) × 10⁷ 

= 3.97298 × 10⁷

b. 1.05 × 10^-4 – 9.7 × 10^-5

= 10.5 × 10^-5 – 9.7 × 10^-5 

= (10.5 – 9.7) × 10^-5

= 0.80 × 10^-5

= 8.0× 10^-6

c. 4.11 × 10^-3 + 8.1 × 10^-4

= 41.1 × 10^-4 + 8.1 × 10^-4

= (41.1 + 8.1) × 10^-4

= 49.2 × 10^-4 

= 4.92 × 10^-3

d. 2.12 × 10⁶ – 3.5 × 10⁵

= 21.2 × 10⁵ – 3.5 × 10⁵

= (21.2 – 3.5) × 10⁵

= 17.7 × 10⁵

= 1.77 × 10⁶

The study of quantitative relations between the amount of reactants and/or products is called stoichiometry.

Sodium hydroxide forms a pale blue precipitate which is insoluble in excess of sodium hydroxide. Ammonium hydroxide forms a pale blue precipitate which redissolves in excess of ammonium hydroxide to form deep blue colouration. 

2Al + 2NaOH + 2H₂O ⟶ 2NaAlO₂ + 3H₂

(Hot and conc.) Sodium meta aluminate

(colourless)

PbO + 2NaOH ⟶ Na₂PbO₂ + H₂O

(Yellow) sodium plumbate

(colourless, soluble)

Aluminium

(i) Hydrogen

(ii) Sodium aluminate, (NaAlO₂ ).

(i) Pb²⁺ ⟶ White soluble in excess

(ii) Fe²⁺ ⟶ Dirty green

(iii) Zn²⁺ ⟶ White soluble in excess

(iv) Fe²⁺ ⟶ Reddish brown

(v) Cu²⁺ ⟶ Blue

(vi) Ca²⁺ ⟶ White insoluble in excess.

 Observations:

(i) White ppt. of Calcium hydroxide is formed.

(ii) Reddish brown ppt. of Fe(OH)₃ is formed.

(iii) Pale blue ppt. is formed which gives deep blue solution with excess of ammonium hydroxide.

(iv) A white ppt. of Lead hydroxide AgCl is formed.

(a) Ferrous salts : Light green

(b) Ammonium salts : Colourless

(c) Cupric salts : Blue

(d) Calcium salts : Colourless

(e) Aluminium salts : Colourless

(i) B and E (Iron III sulphate and magnesium sulphate).

(ii) C and F (Iron III chloride and zinc chloride)

(iii) D (lead nitrate)

(iv) A (copper nitrate)

(v) F (Zinc chloride)

(i) Yellow

(ii) Colourless

(iii) PaleGreen

(iv) Colourless

(v) Colourless