পাতলা H_(2) SO_...

Consider the ionisation of H_(2) SO_(4) as follow" H_(2) SO_(4) + 2H_(2)P rarr 2H_(3) O^(o+) SO_(4)^(2-) The total number of ions furnised by 100 mL of 0.1 M H_(2)SO_(4) will be

Oleum or fuming sulphuric acid contains SO_(3) dissolved in sulphuric acid and has the molecular formula H_(2)S_(2)O_(7) , It is formed by passing SO_(3) in H_(2)SO_(4) . When water is added to oleum, SO_(3) reacts with water to form H_(2)SO_(4) . SO_(3)(g) + H_(2)O(l) to H_(2)SO_(4)(aq) As a result, mass of H_(2)SO_(4) increases. When 100 g sample of oleum is diluted with desired amount of water (in gram) then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling of oleum. % Labelling of oleum = Total mass of H_(2)SO_(4) present in oleum after dilution or = Mass of H_(2)SO_(4) initially present + Mass of H_(2)SO_(4) produced after dilution From this, the percentage composition of H_(2)SO_(4) and SO_(3) (free) and SO_(3) (combined) can be calculated. The percentage of free SO_(3) and H_(2)SO_(4) in 112% H_(2)SO_(4) is

Draw the structures of (i) H_(3)PO_(2) " " (ii) H_(2)SO_(5) " " (iii) H_(2)SO_(3) " " (iv) H_(2)SO_(4) " " (v) H_(2)S_(2)O_(7) .

Balance the following equations: (a) Ca(OH)_(2) + HCI to CaCl_(2) + H_(2)O (b) NaOH+ H_(2)SO_(4) to Na_(2)SO_(4) + H_(2)O © NaCl + H_(2)SO_(4) to NaSO_(4) +HCl (d) Cu + H_(2)SO_(4) to CuSO_(4) + H_(2)O + SO_(2)