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Burette Reading (cm\(^3\)) Rough 1st Titration 2nd Titration 3rd Titration Final Burette Readings (cm\(^3\)) 36.50 30.40 41.70 26.80 Initial Burette readings (cm\(^3\)) 12.30 05.90 17.10 01.90 Vol. of acid used (cm\(^3\)) 24.20 24.50 24.60 24.90 Titre value = \(\frac{24.50 + 24.60}{2}\) = 24.55cm\(^3\) (b)(i) Molar concentration of KMNO\(_4\) \(\frac{C(MnO_4^-) \times V(MnO_4^-)}{C(C_2O_4^{2-}) \times V(C_2O_4^{2-})} = \frac{n(MnO_4^-)}{n(C_2O_4^{2-})}\) \(\frac{C(MnO_4^-) \times 24.55}{0.05 \times 25} = \frac{2}{5}\) C(MnO\(_4^-\)) = \(\frac{0.05 \times 25 \times 2}{24.55 \times 5}\) = 0.0204 mol/dm (ii) Mass concentration of KmnO\(_4\) Molar mass of KmnO\(_4\) = 39 + 55 + (16 x 4) = 158g/mol Mass conc. KmnO\(_4\) = molar conc. x molar mass = 0.0204 x 158 = 3.22g/dm\(^3\) (iii) Volume of CO\(_2\) (gas evolved) \(\frac{n(CO_2)}{n (H_2C_2O_4)}\) = \(\frac{10}{5}\) n(CO\(_2\)) = \(\frac{10}{5}\) x 0.05 = 2 x 0.05 = 0.100 mol V(CO\(_2\)) at s.t.p = n x Vm 25cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) = 0.100 x 22.4dm\(^3\)mol\(^{-1}\) x \(\frac{25}{1000}\) = 0.056dm\(^3\) or 56.0 cm\(^3\) Alternative method Volume of CO\(_2\) gas evolved Amount of H\(_2\)C\(_2\)O\(_4\) used = 0.050 x \(\frac{25}{1000}\) = 0.00125 mol From the equation of the reaction 5 mol of C\(_2\)O\(_4^{2-}\) = 10 mol of CO\(_2\) 0.00125 mol C\(_2\)O\(_4^{2-}\) = 0.00125 mol x \(\frac{10}{5}\) = 0.0025 mol CO\(_2\) @S.T.P 1 mol of CO\(_2\) = 22.4dm\(^3\) 0.00250 mol = 22.4 x 0.0025 = 0.056 dm\(_3\) or 56.0 cm\(^3\)

In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. Cohesive Force is negligible in case of gases.

No expanded working has been attached for this question yet. The verified answer is Option C.

The dissolution of sugar in tea is accompanied by positive change in entropy. When there is a change in the state of a substance from solid to liquid, there is an increase in entropy. In general, if a substance changes its structure from a condensed state to a less condensed state, then we say there is an increase in entropy.

Metal oxides which react with both acids as well as bases to produce salts and water are known as amphoteric oxides. Many metals (such as zinc, tin, lead, aluminium, and beryllium) form amphoteric oxides or hydroxides.

The acetate and paper modify the particulate smoke phase by particle retention (filtration), and finely divided carbon modifies the gaseous phase (adsorption)

Crystallization is required the most in industries where the purity of their products is important e.g in the manufacture of drugs and in sugar production. Options B,C and D do not undergo crystallization

Iron is galvanized with zinc covering to protect it against corrosion. Option A, B, D are the main reasons iron is galvanized

Washing under pressure is an action in the industrial production of H. Options B,C,D will not yield Co and H

The ozone layer or ozone shield refers to a region of Earth's stratosphere that absorbs most of the Sun's ultraviolet (UV) radiation. It contains high concentrations of ozone (O3) relative to other parts of the atmosphere, although still very small relative to other gases in the stratosphere

Vulcanizer is the removal of double bend in a chemical reaction. Option A,B,D can not be removed by vulcanizer

HeSO4 is the acid used in the electrolysis of water. Option A,B,D are concentrated acid and they will react with the water

Sublimation is the transformation of solid gaseous state without passing through the liquid state. Options A,C,D do not show case these character of sublimation as they do not bye pass states. They are in a particular order

Efflorescence is a process of long exposure to the atmosphere which favors salt loss of water. Options B, C, D are not related to loss of water

Ozone means Oz as the atomicity is 3. Option A,B,D cannot give ozone because their atomicity is not 3