Question

Solution:
We know PV = n RT
( 0.9868 times 1=n times 0.082 times 283 )
( mathrm{n}=0.042 mathrm{mol} )
1 mole of Kl reacts with 1 mole of ( mathrm{Cl}_{2} ) to form 1 mole of ( mathrm{I}_{2} )
So 0.042 mole of ( mathrm{KI} ) reacts with 0.042 mole of ( mathrm{Cl}_{2} ) to form
0.042 mole of ( mathrm{I}_{2} )
Amount of ( mathrm{I}_{2} ) liberated ( = ) No. of moles ( mathrm{x} ) molar mass ( =0.042 mathrm{x} )
( 253.81=10.793 mathrm{g} )

# (B) Involving Mass-Volume Relationship 9. What volume of oxygen at 18°C and 750 mm pressure can be obtained from 10 g of potassium chlorate ? 10. What mass of iodine is liberated from a solution of potassium iodide when 1 litre of chlorine gas at 10° C and 750 mm pressure is passed through it?

Solution