he volume of water to be added to 100 cm3 of 5NH.SO. to get decinormal

1.	he volume of water to be added to 100 cm3 of 5NH.SO. to get decinormal concentration is(1) 400 cm(2) 500 cm3(3) 450 cm(4) 100 cm3
Answer» Answer:1)400cm³ Explanation a decinormal concentration would be : N/10 Here, 0.5/10 = 0.05 Converting to N : 0.05×2 = 0.1 formula : N₁V₁ = N₂V₂ Where V is the volume and N the normality. 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

he volume of water to be added to 100 cm3 of 5NH.SO. to get decinormal concentration is(1) 400 cm(2) 500 cm3(3) 450 cm(4) 100 cm3

Answer»

Answer:1)400cm³

Explanation

a decinormal concentration would be :

N/10

Here,

0.5/10 = 0.05

Converting to N :

0.05×2 = 0.1

formula :

N₁V₁ = N₂V₂

Where V is the volume and N the normality.

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