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The volume of water that must be added to a mixture of 250ml of 0.6M HCL and 750ml of 0.2M HCL to obtain 0.25M solution of HCLis:

The volume of water that must be added to a mixture of 250ml of 0.6M HCL and 750ml of 0.2M HCL to obtain 0.25M solution of HCLis:

Grade:11

2 Answers

Gauthami R Menon
19 Points
6 years ago
No. of moles of 0.6M HCl = 0.15
No. of moles of 0.2M HCl = 0.15
let V be the volume(in ml) of the 0.25M solution. No of moles present = 0.15+0.15 = 0.3
\frac{0.3*1000}{V(ml)} = 0.25M
V=1.2L
naman nihal
11 Points
5 years ago
Net moles of solute(HCL in this question) before and after dilution is same
hence MnetVnet = M1V1 + M2V2
V = (M1V1+M2V2)/Mnet
Put the values given in question to get
V net= 1200 ml
now total volume is 1200 ml and intial volume was 1000 ml
hence VOlume of water added is 200 ml

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