Question

900 ml water is added in 100 ml 0.5 M aqueous solution of NaCl. The molarity of resultant solution is​

Answer 1

Answer:

Given that  

Volume of HCl=100 ml=0.1 L

Normality of HCl=0.1 N

Now we can find

Gram-equivalent of HCl=Normality×Volume  

=0.1×0.1=0.01

So,

Volume of solution =Volume of H  2

​  O + Volume of HCl

=900+100=1000 mL

=100 mL=1 L

Normality of  HCl in resulting solution =  1 0.01

​   =0.01 N

pH=−log[H  +  ]

=−log(0.01)

=2        

This is required solution .

Answer 2

Given info : 900 ml water is added in 100 ml , 0.5 M aqueous solution of NaCl.

To find : The molarity of resultant is..

solution : here 900 ml is added in 100 ml, 0.5 M aqueous solution of NaCl.

here you should understand that dilution of solution doesn't change no of moles of solute.

so, initial no of moles of solute = final no of moles of solute.

initial no of moles of solute (i.e., NaCl) = initial volume of solution × initial molarity of solution

= 100ml × 0.5 M

= 100/1000 L × 0.5 M

= 0.05 mol

let resultant molarity of the solution will be x

final no of moles of solute = final volume of solution × final molarity

= (900 ml + 100 ml) × x

= 1000/1000 L × x

= x

∴ x = 0.05

Therefore the molarity of resultant solution is 0.05 M