chemistry

b) From your result and the information provided above calculate CHEMISTRY PRACTICAL ANSWERS 1a ) Solution A = 4.50 gdm ^ -3 of H2 C 2 O4 Solution B = 3.16 gdm ^ -3 of KMnO 4 volume of pipette = 25 .00 cm ^ 3 colour change = orange to pink Equation of the reaction = 4H 2C 2 O4+2 KMnO 4- ---> 8CO 2+ K 2O+MnO3 +4 H2 O TABULATE Burette reading / cm ^ 3 - | Rough | |1st | |2nd | |3 rd| final burette reading - |23 .55 | |46 . 05| | 22. 50 | | 45 .00 | Initial burette reading - |00 .00 | |23 . 55| | 00. 00 | | 22 .50 | Volume of acid used - |23 .55 | |22 . 50| | 22. 50 | | 22 .50 | Average Volume of acide used = 22. 50 + 22 .50 + 22 .50 / 3 = 67. 50 / 3 =22 .50 cm ^ 3 1bi ) Concentration of B in moldm ^ -3 = mass concentration/ molar mass . Where molar mass of KMnO 4 = 39 + 55 + ( 4 * 16 ) = 158 g / mol :concentration in moldm ^ -3 = 316 / 158 = 0. 02mldm ^ -3 1bii ) Concentration of A in moldm ^ -3 CaVa/ CbVb = na / nb Where Ca = concentration of Acide in moldm ^ -3 Cb = concentration of base in moldm ^ -3 Va = volume of acid in Cm^ 3 Vb = volume of base in Cm ^ 3 na= number of mole of acid in the balance equation nb = number of mole of base in the balance equation Making Ca subject of formular Ca = Cb Vb na/ Va nb = 0 .02 25 2/ 22 . 50 1 Ca = 0 .04 Moldm ^ -3 1biii) No of H ^ + in A ldm^ 3 of A = 0 .04 Moldm ^ -3 of H2 C 2O4 H2 C 2O4 dissolved completely in water to give ; H2 C 2O4 ---- --> 2H ^ + C 2 O4^ 2 - 1mole of H2 C 2O4 = 2H ^ + 0. 04moles = 0. 04 * 2 = 0. 05 H^ + but 1mole of H ^ + = 6 .02 * 10 ^ 23 0. 08mole of H ^ + = X X = 0. 08 6.02 10 ^ 23 = 4 .8 * 10 ^ 22 H^ + 1biv) No of K ^ + in B 1mole of substance = 6 .02 * 10 ^ 23 but ; KMnO 4 - ----> K ^ + MnO4 ^ - 1mole of KMnO 4 = K ^ + 0. 02mole of KMnO 4 = 0. 02 * 1 = 0. 02 k^+ but 1mole = 6. 02 * 10^ 23 0. 02mole = 0. 02 6.02 10 ^ 23 = 1.2 * 10 ^ 22 K ^ + 1bv ) PH of A PH of A = Log10 [Moldm ^ 3 of H^ + ] but H 2C 2O4 -- ---> 2H ^ + + C 204 ^ - 1mole of H2 C 2O4 = 2Ca 0. 04mole = Log 10 [2 Ca] = -Log10 [ 2 * 0. 04 ] = -[2 4 10 ^ -2 ] = -[0 .9031 + (-2 Log 10 )] = -[0 .9031 + (-2 * 1)] = -[-0 .9031-2 ) = -[-1 .0969] =1 .1 1biv) POH of B PH + POH = 14 POH = 14 - PH POH = 14 - 1. 1 POH = 12 .9 2) TABULATE. Ca =NaCl and ZnCO3 | TEST | a ) C + water + Filtration. i) C + NaOH in drops , in excess . ii ) C + (aq) NH 3 in drops and in excess . iii ) C + dil HCl in drops , and excess . iv ) C + dil HCl + Bacl 2(aq) v ) C + dil HNO 3 + AgNO3 vi ) C + dil HNO 3 + AgNO3 ^ + excess NH 3 vii) C + few drops of (aq ) NH 3 | OBSERVATION | a ) Dissolves partially to give in black residue and a colourless filterate on filteration i)white gelatinous precipitate formed , soluble in excess NH 3 solution ii ) white gelatinous precipitate formed , soluble in excess iii ) efferrescence due to CO 2 inssuring gas turns moist blue litimus paper red and turns lime water milky iv )white precipitate soluble in dilute HCl v )white preciptate in soluble in dilute HNO 3 acid solution vi )white precipitate dissolved soluble in excess NH 3 vii)No visible reaction | INFERENCE | a ) C is a mixture of soluble and insoluble salts i)Zn ^ 2 +, Pb ^ 2+ or Al ^ 3+ present ii ) Zn +2 + confirmed iii )Gas is CO 2 from CO 3^ 2 - or HCO3^ - iv ) CO 3 ^ 2- confirmed v )CL^ - present vi ) Cl ^ - confirmed vii) Na ^ + present ================================ 3a) The differences in Boiling point of a liquid is (I) they are used to separate the mixture of color (II) with close different boiling point (3b) (i) To make the accurate result of the solution (II) To avoid the error of decimal