3,1 12.The volume of water to be added to 100 cm3 of 0.5 N H2S04 to ge

1.	3,1 12.The volume of water to be added to 100 cm3 of 0.5 N H2S04 to get decinormal concentration isA)100 cm3B) 450 cm3C)500 ст.3D) 400 cm
Answer» Given N as the normality, a decinormal concentration would be : N/10 In this case : 0.5/10 = 0.05 To convert it to N we have : 0.05×2 = 0.1 Using dilution formula : N₁V₁ = N₂V₂ Where V is the volume and N the normality. Doing the substitution: 0.5 × 100 = 0.1 × V₂ V₂ = ( 0.5 × 100) / 0.1 V₂ = 500 cm³ Given that the initial volume that was there is 100 cm³ the volume required is : 500 - 100 = 400 cm³

3,1 12.The volume of water to be added to 100 cm3 of 0.5 N H2S04 to get decinormal concentration isA)100 cm3B) 450 cm3C)500 ст.3D) 400 cm

Answer»

Given N as the normality, a decinormal concentration would be :

N/10

In this case :

0.5/10 = 0.05

To convert it to N we have :

0.05×2 = 0.1

Using dilution formula :

N₁V₁ = N₂V₂

Where V is the volume and N the normality.

Doing the substitution:

0.5 × 100 = 0.1 × V₂

V₂ = ( 0.5 × 100) / 0.1

V₂ = 500 cm³

Given that the initial volume that was there is 100 cm³ the volume required is :

500 - 100 = 400 cm³