DHS_H2_CHEM_P1 ANS

2 BLANK P AGE [Turn Over

3 Section A For each question, there ar e four possible answers A, B, C, and D. Choos e the one you consider to be correct. 1 When 10.0 cm3 of a 0.10 mol dm–3 solution of alkali metal salt MXO3 was reduced with an excess of acidified potassium iodide solution, the resulting iodine required 60.0 cm 3 of 0.10 mol dm –3 sodium thiosulfate solution for its reduction. The anion could be reduced to A XO2 B XO2– C XO– D X– 2 10 cm3 of a hydrocarbon is mixed with 100 cm 3 of oxygen gas which is in excess. The mixture is exploded and after it is cooled to room temperature, the residual gase s occupy a volume of 80 cm 3. Upon passing the gases through potassium hydroxide, th is volume decreases to 50 cm3. What is the formula of the unknown hydrocarbon? A C3H6 B C3H8 C C4H8 D C4H10 Option B: CxHy + (x+y/4) O2  xCO2 + y/2 H2O Volume of CO2 = 30 cm3 By using Avogadro’s Law, Mole ratio  volume ratio  no. of mol of CO2 = 3  x = 3 Volume of O2 remaining = 50 cm3  volume of O2 reacted = 100 – 50 = 50 cm3  x+(y/4) = 50/10 5 = x+(y/4)  y = 8 Hence, formula of the hydrocarbon is C 3H8. Option D: (5-n) e + X5+  Xn+ ---(1) 2I-  I2 + 2e ---(2) I2 + 2S2O3 2-  2I- + S4O6 2- ---(3)  n = -1 No of mol of S2O3 2- = 60.0/1000 x 0.10 = 0.006  No of mol of I = ½ x 0.006 = 0.003 2  no of mol of I- = 0.003 x 2 = 0.006 No of mol of X5+ = 10/1000 x 0.10 = 0.001  6I- = X5+ (2) x 3, -6I  3I + 6e 2  5-n = 6 [Turn Over

4 3 The use of Data Booklet is relevant to this question. The successive ionis ation energies, in kJ mol –1, of an element X are given below. 870 1800 3000 3600 5800 7000 13200 What is element X? A 33As B 53I C 8O D 52Te 4 A 10.0 dm 3 sample of oxygen at a pressure of 250 kPa and 3.0 dm 3 sample of nitrogen at a pressure of 500 kPa are introduced into a 2.5 dm 3 vessel at room temperature. What is the total pressure in the vessel? A 750 kPa B 1200 kPa C 1600 kPa D 1950 kPa Option C: Using Boyle’s Law, 500(3.0) = PO2’ (2.5) P1V1 = P2V2 PN2’ = 600 kPa ’ PO2 VO2 = PO2’ Vvessel  PT = PO2’ + PN2 250(10.0) = PO2’ (2.5) = 1000 + 600 PO2’ = 1000 kPa = 1600 kPa PN2 VN2 = PN2’ Vvessel Option D: The greatest jump is between the 6th and 7th I.E. Hence element X must be in group VI. Therefore, either O or Te is the answer. But with reference to the Data Booklet, the I.E. values do not correspond to that of oxygen. Hence, element X is Te. 5 Given that the standard ent halpy change of combustion of but-