RI Periodic Table II - Answers to Self-Check Questions

-6- A nswers to Self-Check Questions Q1a MgCO3(s)  MgO(s) + CO2(g) Q1b Mg2+ ion has a smaller ionic radius than the Ba 2+ ion. Thus, the Mg 2+ ion has a higher charge density and consequently a stronger polarising power than the Ba 2+ ion. The more polarising Mg2+ ion distorts the electron cloud of the CO 32- anion to a larger extent and hence weakens the covalent bonds within the CO32- anion to a greater extent. Thus, less heat energy is required to decompose MgCO3 than BaCO 3. Hence MgCO 3 decomposes at a lower temperature than BaCO3. Q1 c Mg2+ has the same charge as Ba 2+ but Mg2+ has a smaller ionic radius than Ba 2+. Hence the magnitude of the lattice energy of MgCO3 is larger than that of BaCO3. Q2a MCO3(s)  MO(s) + CO2(g) Mass of CO2 produced = 1.34 g Amount of CO2 = ଵ.ଷସ ସସ.଴ = 0.03045 mol Amount of MCO3 = 0.03045 mol Mass of 0.03045 mol of MCO3 = 4.50 g Mass of 1 mol of MCO3 = 147.8 g Mr of MCO3 = 147.8 Ar of M = 147.8 – 12.0 – (3)(16.0) = 87.8 Hence M is Sr. Q2 b CaMg(CO3)2(s) + 4H+(aq) → Ca2+(aq) + Mg2+(aq) + 2CO2(g) + 2H2O(l) Amount of CO2 = 0.45/44.0 = 0.01023 mol Amount of CaMg(CO3)2 = (½)(0.01023) = 0.005115 mol Mr of CaMg(CO3)2 = 184.4 Mass of CaMg(CO3)2 = 0.005115 x 184.4 = 0.943 g Percentage purity of dolomite = 0.943/1.00 x 100% = 94.3% Q2 c CaSO4.2H2O  CaSO4.nH2O + (2–n) H2O Molar mass of CaSO4.2H2O = 172.2 Amount of CaSO4.2H2O = ଵ଴଴ ଵ଻ଶ.ଶ = 0.5807 mol Mass of H2O lost = 15.7 g Amount of H2O lost = ଵହ.଻ ଵ଼.଴ = 0.8722 mol Molar ratio of CaSO4.2H2O to H2O = 1 : (2–n) = 0.5807 : 0.8722 = 1 : 1.5 Hence (2 – n) = 1.5 n = 0.5 formula of plaster of Paris is CaSO4.½H2O Q2 d(i) Mass of gaseous mixture = 1.00 – 0.438 = 0.562 g Amt of gases = 374/24000 = 0.01558 mol Average molar mass of gaseous mixture = 0.562 / 0.01558 = 36.1 g mol–1 Average Mr = 36.1 Q2 d(ii) The gas which reacted with NaOH(aq) is CO2. Q2 d(iii) Let the other gas be X. Volume of CO2 = 187 cm3 Volume of X = 374 – 187 = 187 cm3 Since volume of X = volume of CO 2, the mixture contains equal amounts of the X and CO2. Mole fraction of each gas = ½ (½)(Mr of X) + (½)(Mr of CO2) = 36.1 (½)(Mr of X) + (½)(44.0) = 36.1 Mr of X = 28.2 Hence X is CO. Q2 d(iv) E is CaC2O4. CaC2O4(s)  CaO(s) + CO(g) + CO2(g) Q3 chlorine bromine iodine Physical state at rtp gas liquid solid Tr end: Volatility decreases in the order: Cl2 > Br2 > I2. Ex planation:  Cl2, Br 2 and I2 exist as simple, non–p olar diatomic molecules with intermolecular instantaneous dipole – induced dipole (id – id) interactions.  Bo iling involves overcoming these intermolecular id– id attractions.  From Cl2 to I2, the size of the electron clou d increases. Thus, the electron cloud becomes more easily distorted/ polarisable,  The id – id attractions become stronger and more energy is required to break the intermolecular forces of attraction heat

-7-  Hence, boiling point increases in the order: Cl2 <