Solution-Class-11-Science-Chemistry-Chapter-2-Introduction to Analytical Chemistry-Maharashtra Board

(d) Analytical chemistry

(c) Colour

(a) kg

(b) 3

(b) 4

(c) 14.0 g

(b) 27

(d) A₂B₃

(c) CH₃

(c) 24

(b) 36 %

(b) Plotting a graph

Least count : It is the smallest quantity or value that can be measured on the measuring equipment or instrument.

Significant figures : The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.

Rules for deciding significant figures :

• All nonzero digits are significant. For example 127.34 g contains five significant figures which are 1, 2, 7, 3 and 4.
• All zeros become significant, if they are placed between two non zero digits. For example, 120.007 m contains six significant figures which are 1, 2, 0, 0, 0 and 7.
• Zeros on the left of the first non zero digit are not significant. Such a zero indicates the position of the decimal point. For example, 0.025 has two significant figures,
• 2 and 5. 0.005 has only one significant figure, 5.
• All zeros placed at the end of a number are significant if they are on the right side of the decimal point. They represent the accuracy or precision of the measuring scale.
• For example, 243.0 has four significant figures which are 2, 4, 3 and 0.
• Terminal zeros are not significant if there is no decimal point. For example, the measurement 400 g has one significant figure which is 4.
• All digits in numbers written in scientific notation are significant. For example, 2.035 x 10² has four significant figures which are 2, 0, 3 and 5.
• The measurement, 3.25 x 10^-5 has three significant figures viz. 3, 2 and 5.

Percentage composition : The percentage by mass of each element in a compound is called per cent composition.

Empirical formula : The simplest ratio of atoms of the constituent elements in a molecule is called the empirical formula of that compound.

Molecular formula : Molecular formula of a compound is the formula which indicates the actual number of atoms of the constituent elements in a molecule.

Limiting reagent : The reactant in a chemical reaction that limits the amount of product that can be formed is called the limiting agent. The reaction will stop when all of the limiting reactant is consumed.

• A balanced chemical equation gives the ideal stoichiometric relationship among reactants and products. However, when a chemist carries out a reaction, reactants for the experiment are not necessarily present in exact stoichiometric amounts. This is because, the goal of the reaction is to produce the maximum quantity of a useful compound or product from the starting material.
• Many a times, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted into the desired product. Thus, the reactant which is present in lesser amounts gets consumed after some time and subsequently, no further reaction takes place.

In 1960, the General Conference of weight and measures adopted an International System of Units, called SI units (French name System internationale d'Unites) which has seven fundamental units.

The SI unit of mass is kilogram (kg).

Mole fraction (x) : The mole fraction of any component of a solution is defined as the ratio of number of moles of that component present in the solution to the total number of moles of all the components of the solution.

Mole fraction = \(\frac{\text{Number of moles of component}}{\text{Total no of moles of solute and solvent in solution}}\)

Number of moles of a substance = \(\frac{\text{Mass of substance}}{\text{Molar mass of substance}}=\frac{W}{M}\) mol

If a solution contains n₁ moles of solvent and n₂ moles of solute, then the total number of moles present in the solution = n₁ + n₂.

Mole fraction of the solvent x₁ = \(\frac{n_1}{n_1+n_2}\)

Mole fraction of the solute x₂ = \(\frac{n_2}{n_1+n_2}\)

Mole fraction is independent of temperature and has no units.

Molarity (M) : It is defined as the number of moles of a solute present in 1 litre of a solution.

If n moles of a solute are present in V lit. of a solution then,

Volume of solution in lit.

Molarity(M) = \(\frac{\text{Number of moles of solute}}{\text{Volume of solution in lit}}=\frac{n}{V}\)  mol L^-1

Molarity = \(\frac{\text{Mass of solute}}{\text{Molar mass of solute × Volume of solution in L.}}\)

= \(\frac{\text{Mass of solute×1000}}{\text{Molar mass of solute × Volume of solution in}cm^3}\)

The unit of molarity is mol litre^-1 (mol L^-1)

Molarity changes with temperature.

Molality (m) ; It is defined as the number moles of a solute dissolved in 1 kg (or 1000 gram) of a solvent.

Molality (m) = \(\frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}\)

= \(\frac{\text{Mass of solute}}{\text{Molar mass of solute × Mass of solvent in kg}}\)

= \(\frac{\text{Mass of solute × 1000}}{\text{Molar mass of solute × Mass of solvent in g}}\)

The unit of molality is mol kg^-1 and denoted by m.

Molality is independent of temperature, since it involves mass of solute and mass of solvent which are independent of temperature.

Stoichiometry : The study of quantitative relations between the amount of reactants and products is called stoichiometry.

The final result of a calculation often contains figures that are/not significant. When this occurs the final result is rounded off.

Molarity of a solution depends upon temperature since it involves volume of the solution which is temperature dependent.

Analytical chemistry : 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.

Importance of accurate measurements :

• Accurate measurements are important because exact amounts of reagents or chemicals are required for reactions to take place.
• Measurements that are not accurate, provide incorrect data which can lead to wrong or even dangerous conclusions or results.
• For example, if a lab experiment requires a specific amount of a chemical, measuring the wrong amount may result in an unsafe or unexpected outcome.
• Due to all these reasons, accurate measurement is crucial in science.

(a) 4, (b) 3, (c) 2, (d) 5, (e) 3, (f) 2, (g) 3, (h) 4

(a) The last two digits being 55, we first convert first two digits from 25 to 26. For two significant figures the answer is 26 mL.

(b) Since the third digit 4 is less than 5 and we require two significant figures, we drop the third digit 4 and write 2.5 x 10^-3 m

(c) 1.5 x 10⁵ mg

(d) 2.0 x 10² g

(a) 1.43 cm³

(b) 458 x 10² cm

(c) 643 cm²

(d) 3.90 x 10^-2 m

(e) 6.40 x 10^-3 km : In the figure 6.398 × 10^-3  the fourth digit is more than 5, hence third digit is increased by 1 which makes second digit 4 and final figure with three significant figures becomes 6.40 x 10^-3 km.

(f) 1.79 x10^-2 g

(g) 790 x 10² or 7.90 x 10⁴ m

(h) 4.22 x 10⁴

(i) 650

(j) 2.36 x 10⁷ mm

(k) 4.20 x 10^-3 kg

(a) 2.3 × 10³ mL +4.22 × 10⁴ mL + 9.04 × 10³ mL + 8.71 x 10⁵ mL

= (0.23 + 4.22 + 0.904 + 87.1) × 10⁴

= 92.454 × 10⁴ mL

= 92.5 × 10⁴ mL

= 9.2 × 10⁵ mL

(b) 319.5 g - 20460 g - 0.0639 g - 45.642 g - 4.173 g

= - 20190.38 g

(a) 3.498 × 10^-4

(b) 2.354678 × 10²

(c) 7.00000 × 10⁴

(d) 1.56900 × 10³

(a) 4, (b) 4, (c) 3, (d) 3

(a) 12300

(b) 2.030 × 10^-3

(c) 12300

(d) 0.000189

(a) Molar mass of Copper sulphate crystal (CuSO₄.5H₂O)

= Atomic mass of Cu + Atomic mass of S + 4 × Atomic mass of O + 5(2 × Atomic mass of H + Atomic mass of O)

=63.5+32+4×16+5(2×1+16)

=63.5+32+64+5(18)

=249.5 g/mol

(b) Molar mass of Sodium carbonate, decahydrate (Na2CO₃.10H₂O)

= 2 × Atomic mass of Na + Atomic mass of C + 3 × Atomic mass of O + 10(2 × Atomic mass of H + Atomic mass of O)

=2×23+12+3×16+1O (2×1+16)

=46+12+48+10(18) .

=286g/mol

(c) Molar mass of Mohr's salt [FeSO₄(NH₄)₂SO₄.6H₂O]

= Atomic mass of Fe + Atomic mass of S + 4 × Atomic mass of O + 2(Atomic mass of N + 4 × Atomic mass of H) + Atomic mass of S + 4 × Atomic mass of O + 6(2 x Atomic mass of H + Atomic mass of O)

=56+32+4×16+2(14+4×1)+32+4 x 16+6 (2×1+16)

=56+32+64+2(18)+32+64+6 (18)

= 392 g/mol

(a) Percentage composition of the constituent elements in [Pb₃(PO₄)₂]

Molecular mass of Lead phosphate [Pb₃(PO₄)₂]

= 3 x Atomic mass of Pb + 2 (Atomic mass of P + 4 x Atomic mass of O)

=3 x 207+2 (31+4x16) = 621+2(31+64)

= 811 g/mol

(1) To determine the percentage composition of Pb

Mass of Pb in Pb₃(PO₄)₂ = 3 x 207 = 621

= \(\frac{\text{Mass of Pb in}Pb_3 (PO_4)_2}{\text{Molecular mass of}Pb_3 (PO_4)_2}\)  x 100%

= \(\frac{621}{811}\) x 100% = 76.57%

(2) To determine the percentage composition of P

= Mass of P in Pb₃(PO₄)₂ = 2 x 31 = 62

= \(\frac{\text{Mass of P in}Pb_3 (PO_4)_2}{\text{Molecular mass of}Pb_3 (PO_4)_2}\)  x 100%

= \(\frac{62}{811}\)x 100% = 7.64%

(3) To determine the percentage composition of O

= Mass of O in Pb3(PO4)2 = 2 (4 x 16) = 128

= \(\frac{\text{Mass of O in}Pb_3 (PO_4)_2}{\text{Molecular mass of}Pb_3 (PO_4)_2}\) x 100

= \(\frac{128}{811}\) x 100% = 15.78%

(b) Percentage composition of the constituent elements in [K₂Cr₂O₇]

Molecular mass of K₂Cr₂O₇

= 2 x Atomic mass of K + 2 x Atomic mass of Cr + 7 x Atomic mass of O

= 2x39+2x52+7x16 = 294

(1) To determine the percentage composition of K

= Mass of K in K₂Cr₂O₇ = 2 x 39 = 78

= \(\frac{\text{Mass of k in}K_2Cr_2O_7}{\text{Molecular mass of}K_2Cr_2O_7}\) x 100%

= \(\frac{78}{294}\) x 100% = 26.53%

(2) To determine the percentage composition of Cr

= Mass of Cr in K₂Cr₂O₇ = 2 x 52 = 104

= \(\frac{\text{Mass of Cr in}K_2Cr_2O_7}{\text{Molecular mass of}K_2Cr_2O_7}\)x 100%

= \(\frac{104}{294}\)x 100% = 35.37%

(3) To determine the percentage composition of O

= Mass of O in K₂Cr₂O₇ = 7 x 16 = 112

= \(\frac{\text{Mass of O in}K_2Cr_2O_7}{\text{Molecular mass of}K_2Cr_2O_7}\) x 100%

= \(\frac{112}{294}\) x 100% = 38.10%

(c) Percentage composition of the constituent elements in (NaNH₄HPO₄.4H₂O)

Molecular mass of NaNH₄HPO₄.4H₂O

= Atomic mass of Na + Atomic mass of N + 4 x Atomic mass of H + Atomic mass of H + Atomic mass of P + 4 x Atomic mass of O + 4(2 x Atomic mass of  H + Atomic mass of O)

= 23+14+4x1+1+31+4x16+4 (2x1+16)

= 209 g/mol

(1) To determine the percentage composition of Na

= Mass of Na in NaNH₄HPO₄.4H₂O = 23

= \(\frac{\text{Mass of Na in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\) x 100%

= \(\frac{23}{209}\) x 100% = 11.00%

(2) To determine the percentage composition of N

= Mass of N in NaNH₄HPO₄.4H₂O = 14

= \(\frac{\text{Mass of N in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\) x 100%

= \(\frac{14}{209}\) x 100% = 6.70%

(3) To determine the percentage composition of H

= Mass of H in NaNH₄HPO₄.4H₂O = 13x1=13

= \(\frac{\text{Mass of H in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\) x 100%

= \(\frac{13}{209}\)x 100% = 6.22%

(4) To determine the percentage composition of P

= Mass of P in NaNH₄HPO₄.4H₂O = 31

= \(\frac{\text{Mass of P in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\)x 100%

= \(\frac{31}{209}\)x 100% = 14.83%

(5) To determine the percentage composition of O

= Mass of O in NaNH₄HPO₄.4H₂O = 8x16 = 128

= \(\frac{\text{Mass of O in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\)x 100%

= \(\frac{128}{209}\)x 100% = 61.24%

(A) Percentage composition of the constituent elements in FeSO₄.7H₂O

Molecular mass of FeSO₄.7H₂O

= Atomic mass of Fe + Atomic mass of S + 4 x Atomic mass of O + 7(2 x Atomic mass of H + Atomic mass of O)

=56+32+4x16+7(2x1+16)

= 278 g/mol

(1) To determine the percentage composition of Fe

= Mass of Fe in FeSO₄.7H₂O = 56

= \(\frac{\text{Mass of Fe in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\) x 100%

= \(\frac{56}{278}\) x 100% = 20.14%

(2) To determine the percentage composition of O

= Mass of O in FeSO₄.7H₂O = 11x16 = 176

= \(\frac{\text{Mass of O in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\)x 100%

= \(\frac{176}{278}\)x 100% = 63.31%

(3) To determine the percentage composition of S

= Mass of S in FeSO₄.7H₂O = 32

= \(\frac{\text{Mass of S in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\)x 100%

= \(\frac{32}{278}\)x 100% = 11.51%

(4) To determine the percentage composition of H

= Mass of H in FeSO₄.7H₂O = 14x1=14

= \(\frac{\text{Mass of H in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\)x 100%

= \(\frac{14}{278}\)x 100% = 5.04%

To find out the mass of iron and the water of crystallisation in 4.54 kg of the crystals.

Weight of crystals = 4.54 kg

Mass of iron = ?

278 g of FeSO₄.7H₂O contains 56 g of Fe,

∴ 4.54 kg of FeSO4.7H2O contains

\(\frac{4,54×56}{278}\) = 0.915 kg of Fe

Mass of iron in 4.54 kg = 0.915 kg

Mass of water of crystallisation = ?

278 g of FeSO₄.7H₂O contains 7H₂O molecules

i.e 7 x 18 g of H₂O= 126 g of H2O

∴ 4.54 kg of FeSO4.7H2O contains

\(\frac{4,54×126}{278}\) = 2.058 kg of water of crystallisation

Mass of water of crystallisation = 2.058 kg

Given :

Atomic mass of Fe = 55.84

Percentage of iron in haemoglobin = 0.335 %

One molecule of haemoglobin contains 4 atoms of iron.

Since, 1 atom of Fe = 55.84 g

4 atoms of Fe = 223.36

0.335 g of Fe is present in 100 g of haemoglobin

223.36 g of Fe is present in

=\(\frac{223.36×100}{0.335}\)  = 66674.6 g/mol of haemoglobin

Molecular weight of haemoglobin = 66674.6 g/mol

Given : Na = 43.4 %, C = 11.3 % and O = 45.3 %.

To calculate the moles of the constituent elements

Moles of Na = \(\frac{\text{% of Na}}{\text{Atomic mass of Na}}=\frac{43.4}{23}\)  = 1.89

Moles of C = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{11.3}{12}\)  = 0.941

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{45.3}{16}\)  = 2.83

Hence, the ratio of number of moles of Na : C : O is

\(\frac{1.89}{0.941}\)  = 2 : \(\frac{0.941}{0.941}\)   = 1 : \(\frac{2.83}{0.941}\)   = 3

Hence the empirical formula is Na₂CO₃

Given :

Atomic weight of M = 56

Percentage of M = 70.0 %

Since, the oxide contains 70.0% of M, the amount of oxygen will be 30.0 %

The atomic weight of Oxygen = 16

Moles of M = \(\frac{\text{% of Na}}{\text{Atomic mass of M}}=\frac{70}{56}\) =  1.25

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{30}{16}\) = 1.875

Hence, the ratio of the number of moles of the metal with oxygen would be,

\(\frac{1.25}{1.25}\) = 1 and \(\frac{1.875}{1.25}\)= 1.5

i.e. 2 : 3

Hence, the empirical formula of the compound would be M₂O₃

Given :

Iron = 0.2014 g, Sulphur = 0.1153 g, Oxygen = 0.2301 g,

Water of crystallisation = 0.4532 g, Wt. of hydrated salt: 1.00 g

To calculate the percentage of the different elements.

Percentage of Fe in the salt = \(\frac{\text{Wt of Fe}}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.2014}{1.00}\) x 100 =  20.14%

Percentage of S in the salt = \(\frac{\text{Wt of S}}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.1153}{1.00}\) x 100 =  11.53%

Percentage of O in the salt = \(\frac{\text{Wt of O}}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.2301}{1.00}\) x 100 = 23.01%

Percentage of H₂O in the salt = \(\frac{\text{Wt of }H_2O}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.4532}{1.00}\) x 100 = 45.32%

To calculate the moles of each element

Moles of Fe  =\(\frac{\text{% of Fe}}{\text{Atomic mass of Fe}}=\frac{20.14}{56}\)  = 0.360

Moles of S  =\(\frac{\text{% of S}}{\text{Atomic mass of S}}=\frac{11.53}{32}\)  = 0.360

Moles of O  = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{23.01}{16}\)  = 1.438

Moles of H₂O  = \(\frac{\text{% of }H_2O}{\text{Atomic mass of} H_2O}=\frac{45.32}{18}\) = 2.518

Hence, the ratio of number of moles of Fe : S : O : H₂O

Is \(\frac{0.360}{0.360}\) = 1 : \(\frac{0.360}{0.360}\) = 1 : \(\frac{1.438}{0.360}\) = 3.99 : \(\frac{2.518}{0.360}\) = 7 :

Hence, the empirical formula is FeSO₄.7H₂O

Given :

Carbon = 20 %, Hydrogen = 6.7 %, Nitrogen = 46.67 % Molecular mass = 60

The total percentage of the constituent elements

= % Carbon + % Hydrogen + % Nitrogen

= 20 + 6.7 + 46.67

= 73.37 %

This is less than 100%.

Hence, the compound contains adequate oxygen so that the total percentage of elements is 100 %.

Hence, the percentage of oxygen = 100 — 73.37 = 26.63 %

To calculate the moles of the constituent elements

Moles of C  = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{20}{12}\)  = 1.66

Moles of H  = \(\frac{\text{% of H}}{\text{Atomic mass of H}}=\frac{6.7}{1}\) = 6.7

Moles of N  = \(\frac{\text{% of N}}{\text{Atomic mass of N}}=\frac{46.67}{14}\) = 3.33

Moles of O  = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{26.63}{16}\) = 1.66

Hence, the ratio of the number of moles of

C : H : N : O is

\(\frac{1.66}{1.66}\) = 1 : \(\frac{6.7}{1.66}\) = 4.03 : \(\frac{3.33}{1.66}\)  = 2.00 : \(\frac{1.66}{1.66}\)= 1 :

Hence, the empirical formula is CH₄N₂O

Empirical formula mass

= Atomic mass of C + 4 x Atomic mass of H + 2 x Atomic mass of N + Atomic mass of O

=12+4x1+2x14+16 = 60

Molar mass = Empirical mass (Since molar mass = Molecular mass)

Molecular formula = Empirical formula CH₄N₂O

Given :

Carbon = 54.55, % Hydrogen = 9.09 %, Oxygen = 3636 %, Molecular mass = 88

Moles of C  = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{54.55}{12}\)  = 4.54

Moles of H  = \(\frac{\text{% of H}}{\text{Atomic mass of H}}=\frac{9.09}{1}\)  = 9.09

Moles of O  = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{36.36}{16}\) = 2.27

Hence, the ratio of the number of moles of C : H : O is

\(\frac{4.54}{2.27}\) = 2 : \(\frac{9.09}{2.27}\) = 4 : \(\frac{2.27}{2.27}\) = 1

Hence, the empirical formula is C₂H₄O

Empirical formula mass = 2 x Atomic mass of C + 4 x Atomic mass of H + Atomic mass of O = 2x12+4x1+16 = 44

r = (\frac{\text{Molar mass}}{\text{Emperical formula mass}}\)=\(\frac{88}{44}\) = 2

Molecular formula = r x Empirical formula

Molecular formula = 2 x C₂H₄O = C₄H₈O₂

Given :

Amount of carbohydrate = 310 g

Amount of Carbon residue obtained on heating = 124 g

Amount of H₂O formed = Amount of carbohydrate - Amount of carbon as residue

= 310 − 124 = 186 g

Since 18 g of H₂O contains 2 g of H

186 g of H₂O will contain \(\frac{186×2}{18}\) = 20.66 g of H

Since 18 g of H₂O contains 16 g of O

186 g of H₂O will contain \(\frac{186×16}{18}\) = 165.33g of O

To determine the moles of the C, H and O in the compound.

Moles of C  = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{124}{12}\)  = 10.33

Moles of H  = \(\frac{\text{% of H}}{\text{Atomic mass of H}}=\frac{20.66}{1}\) = 20.66

Moles of O  = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{165.33}{16}\)  = 10.33

Hence, the ratio of the number of moles of C : H : O is

\(\frac{10.33}{10.33}\) = 1 : \(\frac{20.66}{10.33}\) = 2 : \(\frac{10.33}{10.33}\) = 1

Hence, the empirical formula is CH₂O

(a) 3.672199 x 10⁶

(b) 9.8 x 10^-5

(c) 4.61 x 10^-3  

(d) 1.9875 x 10²  

(a) 0.0000000000349

(b) 0.375

(c) 51600

(d) 0.004371

(e) 0.000011

(f) 0.143

(g) 477

(h) 5.00858585

(a) \(\frac{1}{3.40×10^{24}}\)= 0.2941 x 10^-24 = 2.9 x 10^-25   

(b) \(\frac{33}{9.00×10^{−4}}\)= 3.66 x 10⁴ = 3.7 x 10⁴   

(c) \(\frac{1.4×10^9}{(2.77×10^3)(3.76×10^5}\) = \(\frac{1.4×10^9}{10.4152×10^8}\) 

= 0.1344 x 10 = 1.344 = 1.3   

(d) \(\frac{(4×10^{-3})(9.9×10^{-7})}{(789)(1.002×10^{-10})(0.3×10^2}\) = 0.169 x 10^-2 = 1.7 x 10^-3   

(a) (3.26 × 10⁴) (1.54 × 10⁶) = 5.0204 × 10¹⁰ = 5.02 × 10¹⁰

(b) (8.39 × 10⁷) (4.53 × 10⁹) = 38.0067 × 10¹⁶ = 3.80 × 10¹⁷

(c) \(\frac{8.94×10^6}{4.35×10^4}\) = 2.055 × 10² = 2.06 × 10²

(d) \(\frac{(9.28×10^9)(9.9×10^{-7})}{(511)(2.98×10^{-6})}\) = \(\frac{91.872×10^2}{1522.78×10^{-6}}\)

= 0.06033 × 10⁸ = 6.03 x 10⁶

(a) 3.971 × 10⁷ + 1.93 × 10⁴

=3971 × 10⁴ + 1.93 × 10⁴

= (3971 + 1.93) × 10⁴

= 3972.98 × 104

= 3.97293 × 10⁷

(b) 1.05 × 10⁻⁴ − 9.7 × 10⁻⁵

= 1.05 ×10−4 − 0.97 × 10−4

= (1.05 − 0.97) × 10⁻⁴

= 0.08 × 10⁻⁴ = 8.0 × 10⁻⁶

(c) 4.11 × 10⁻³ + 8.1 × 10⁻⁴

= 4.11 × 10⁻³ +0.81 × 10⁻³

= (4.11 + 0.81) × 10⁻³

= 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

Given :

Volume of acetone taken = 1.000 mL

Mass of bottle = 38.0015 g

Actual mass of acetone = 0.7791 g

To determine the mean values of the weight of acetone (Observed value)

Mean values of the weight of acetone =\(\frac{0.7783+0.7780+0.7786}{3}\)  = 0.7783 g

Absolute error = Observed value — True value

= 0.7783 − 0.7791 = − 0.008

Relative error = \(\frac{\text{Absolute error}}{\text{True value}}\) x 100%

= \(\frac{0.008}{0.7791}\) = 0.1027%

Accuracy of the measurements is 0.1027 %.

To determine the average deviation of the measurements :

Mean absolute Deviation = \(\frac{0.0000+0.0003+0.0003}{3}\)  = ± 0.0002

Relative deviation = \(\frac{\text{Absolute error}}{\text{True value}}\)  x 100%

= \(\frac{0.0002}{0.7783}\)  x 100% = 0.0257 %

Relative error = 0.1027%

Precision of the measurements = 0.0257%

The measurement 12.6 is acceptable.

Reason :

(a) 5 inches is approximately 12.7 cm. The value 12.6 is closer to this value, hence this measurement is acceptable.

(b) Since the length of the box has been measured using a metre stick graduated in millimeters.

The least count of the scale is 0.1 cm. The other measurements cannot be obtained using a scale with a least count of 0.1 cm, hence the other measurements can be rejected.

Given :

Weight of calcium carbonate = 19.3 g

The balanced chemical equation is,

CaCO_3(s) > CaO_(g) + CO_2(g)

Since 100 g of CaCO₃ produces 56 g of CaO

19.3 g of CaCO₃ will produce

 \(\frac{19.3×56}{100}\) = 10.8 g of caO

The balanced chemical equation for the burning of sucrose is,

C₁₂H₂₂O_11(s) + 12O_2(g) —> 12CO_2(g) + 11H₂O_(l)

342g          12 x (32)

= 384g

342 g of sucrose requires 384 g of oxygen

 34 g of sucrose will require

 \(\frac{34×384}{342}\) = 38.1754 g of oxygen

The hourly requirement of oxygen is 38.175 g

Therefore, the requirement of oxygen for one day will be = 38.1754 x 24

= 916.209 g

= 916.21g

The oxygen requirement for one day = 916.21 g.