(1) 1 M(2) 0.5M(3) 2MThe volume of water that must be added to a mixtu

1.	(1) 1 M(2) 0.5M(3) 2MThe volume of water that must be added to a mixture of 250 ml of 06 M HCl and em0.2 M HCI to obtain 0.25 M solution of HCl is!(4) 300 m
Answer» No. of moles of 0.6M HCl = 0.15 No. of moles of 0.2M HCl = 0.15 letVbe the volume(in ml) of the 0.25M solution. No of moles present = 0.15+0.15 = 0.3 0.25M=(0.3×1000)/ VV=1.2L=1200 ml now total volume is 1200 ml and intial volume was 1000 ml Hence volume of water added is 200 ml

(1) 1 M(2) 0.5M(3) 2MThe volume of water that must be added to a mixture of 250 ml of 06 M HCl and em0.2 M HCI to obtain 0.25 M solution of HCl is!(4) 300 m

Answer»

No. of moles of 0.6M HCl = 0.15

No. of moles of 0.2M HCl = 0.15

letVbe the volume(in ml) of the 0.25M solution. No of moles present = 0.15+0.15 = 0.3

0.25M=(0.3×1000)/ VV=1.2L=1200 ml

now total volume is 1200 ml and intial volume was 1000 ml

Hence volume of water added is 200 ml