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First Year Chemistry Solutions


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Class 9Class 10First YearSecond Year
Whatismolarityandgiveitsformulatopreparemolarsolution?What is molarity and give its formula to prepare molar solution?

Worked Example 2: The pH of a solution is 9.63. Compute its hydrogen ion concentration.
Worked Example 2: The pH of a solution is 9.63. Compute its hydrogen ion concentration.

WorkedExample2:ThepHofasolutionis9.63.Computeitshydrogenionconcentration.Worked Example 2: The pH of a solution is 9.63. Compute its hydrogen ion concentration.

(b) Give the oxidation number of:(vi) \mathrm{O} in \mathrm{OF}_{2}
(b) Give the oxidation number of:(vi)  \mathrm{O}  in  \mathrm{OF}_{2}

(b)Givetheoxidationnumberof:(vi)OinOF2(b) Give the oxidation number of:(vi) \mathrm{O} in \mathrm{OF}_{2}

Example 2: The oxidation of \mathrm{FeSO}_{4} by \mathrm{KMnO}_{4} in acidic solution
Example 2: The oxidation of  \mathrm{FeSO}_{4}  by  \mathrm{KMnO}_{4}  in acidic solution

Example2:TheoxidationofFeSO4byKMnO4inacidicsolutionExample 2: The oxidation of \mathrm{FeSO}_{4} by \mathrm{KMnO}_{4} in acidic solution

10. Tyndall effect is due to:(a) blockage of beam of light(b) non-scattering of beam of light(c) scattering of beam of light(d) passing through beam of light
10. Tyndall effect is due to:(a) blockage of beam of light(b) non-scattering of beam of light(c) scattering of beam of light(d) passing through beam of light

10.Tyndalleffectisdueto:(a)blockageofbeamoflight(b)nonscatteringofbeamoflight(c)scatteringofbeamoflight(d)passingthroughbeamoflight10. Tyndall effect is due to:(a) blockage of beam of light(b) non-scattering of beam of light(c) scattering of beam of light(d) passing through beam of light

6. A hiker walks due east at 4 \mathrm{~km} per hour and a second hiker starting at the same point walks 55^{\circ} north-east at the rate of 5 \mathrm{~km} per hour. How far apart will they be after 3 hours?
6. A hiker walks due east at  4 \mathrm{~km}  per hour and a second hiker starting at the same point walks  55^{\circ}  north-east at the rate of  5 \mathrm{~km}  per hour. How far apart will they be after 3 hours?

6.Ahikerwalksdueeastat4 kmperhourandasecondhikerstartingatthesamepointwalks55northeastattherateof5 kmperhour.Howfarapartwilltheybeafter3hours?6. A hiker walks due east at 4 \mathrm{~km} per hour and a second hiker starting at the same point walks 55^{\circ} north-east at the rate of 5 \mathrm{~km} per hour. How far apart will they be after 3 hours?

\mathrm{Q} 21.4 .675 \mathrm{~g} of a compound with empirical formula \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O} were dissolved in 212.5 \mathrm{~g} of pure benzene. The freezing point of solution was found 1.02^{\circ} \mathrm{C} less than that of pure benzene. The molal freezing point constant of benzene is 5.1^{\circ} \mathrm{C} . Calculate (i) the relative molar mass and (ii) the molecular formula of the compound.
 \mathrm{Q} 21.4 .675 \mathrm{~g}  of a compound with empirical formula  \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}  were dissolved in  212.5 \mathrm{~g}  of pure benzene. The freezing point of solution was found  1.02^{\circ} \mathrm{C}  less than that of pure benzene. The molal freezing point constant of benzene is  5.1^{\circ} \mathrm{C} . Calculate (i) the relative molar mass and (ii) the molecular formula of the compound.

Q21.4.675 gofacompoundwithempiricalformulaC3H3Oweredissolvedin212.5 gofpurebenzene.Thefreezingpointofsolutionwasfound1.02Clessthanthatofpurebenzene.Themolalfreezingpointconstantofbenzeneis5.1C.Calculate(i)therelativemolarmassand(ii)themolecularformulaofthecompound. \mathrm{Q} 21.4 .675 \mathrm{~g} of a compound with empirical formula \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O} were dissolved in 212.5 \mathrm{~g} of pure benzene. The freezing point of solution was found 1.02^{\circ} \mathrm{C} less than that of pure benzene. The molal freezing point constant of benzene is 5.1^{\circ} \mathrm{C} . Calculate (i) the relative molar mass and (ii) the molecular formula of the compound.

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