Question

10 ml of N/2 HCL, 20 ml of M / 5 H2SO4 , 20 ml of N/4 H3PO4 are mixed together. the volume of water that must be added to make a 0.18 N solution is? ( assume there is no change in volume on mixing). Ans.50

EXPLAINATION REQUIRED!!!!!!!

Answer 1

Answer:

Meq= N × V

where Meq is the equivalent molarity

N is normality of the solution

V is volume of solution

Applying the formula to each of the solution separately

1) HCl

Meq= 1× 5 = 5M

2) H2SO4

Meq=

2

1

×20=10 M

3) HNO3

Meq=

3

1

×30=10 M

For total solution

Meq = M1 + M2 + M3

Meq = 5+10+10 = 25 M

and volume = 1000 mL

Thus applying the same formula

Meq= N × V

25 = N× 1000

N= $$\dfrac{25}{1000} N

=

40

N

Thus normality of the solution would be N/40

.

Answer 2

Answer:

Explanation:

No. of meq of  H += 10 *1 +  20*2 =50           (h2so4  =N= 2m)    

No. of meqof  O H   -=30×1=30

No. of meq of  H+   left unreacted = 50 - 30 = 20 meq  

Hence, (A) is correct , (