Question

1. Determine whether each of the following will undergo dissociation, dissolving or ionization. Then write the solvation equation for each substance: a) Lithium oxide - b) Magnesium nitride - Mga Na c) Hydrobromic acid

Answer 1

1. (a) Lithium oxide- ligo has very high Hence, it binding energy. shall undergo ionisation . ionisation Ligo alito? (b) Magnesium nitride - Here Mg ₂ N2 can loose a nitrogen gas molecule, which is thermodynamically favoured. Hence, it shall undergo dissociation as : Mg ₂ Ngo 3mg + N2 P acid) (6) HBr (Hydrobromic so contains can a release proton (H+) and bromide (Br-) in water, both of which undergo dissolving can Hence, it shall be solvated.

4. Molecular wt of potassium nitrate (KNO3): 101.10 gmot Thus; 0.225 g of KNO3 = 0.225 moles 101 10 0.0022 moles Thus ; on 750 ml, moles = 0.0022 1000 mL, moles: 0.0022 600x 1000 moles " In = 0.029 (M) is strength = 0.029 (M) 5. Mol wt of H₂50g = 98 g on 1000 m 2 (M) Hasou ; wt required = 2x98 g ign 750 mi 2(M) Hason; " " 2x98x750 . cone = 147 g Precautions: 147 g of H₂sou is to be added very slowly through a funnel to a chilled beaker water glass may otherwise, the containing crack 1000

b. Initially, on 1000 In 750 mot.ml, ml, moles of " " Nad = 0.330 ) 11: 0.330 x 750 1000 When the student adds 1.15 L of water, total volume = 11150+ 750) mL = 1900 mL of Nad = . In 1900 mi, no of moles ... 4 0.2475 0.2495 * 1000 • 9n 1000 ML 1900 0.130 (H) Visi & V2 32 ] 1 7. det volume be v mL . V x 10.0 = 0.5 x 2500 on ve 0.5 x 2500 ml 10 v= 125 mL

- Pbal2 0.223 pbet + 2 a 0.223 0.446 concentration 1 1 of Pb2+ = 0.283.(M) 0.446.(M) 9. det, after first dilution, strength be S, (M) i. Sex (1350+15) = 0.5X15 ori si : 158015 1 13655 ans Sc4ax 10-9 (N) det concentration be S2 CM) after second dilution le Sy(1365+1350) = 5.49 x1013 x 1365 om, sa 2.968 1019 (M)

. 10. (a) Total volume : 600 mL a strength of AIB Ng = 10 boo = 0.2 (M) & of concentration S Al3+ = 0.2 (M) By' 0:6(M) i billa x 1000 (M) 16) Molarity of Nael - - 38.5 250 = 0.437 (M) i cone of wat = cone of t = 0.437 IM) Molarity of Kl = sinh 1 X 100 CM) - 0.279 (M) cone of k t = one of a = 0.279(M)