Question

give steps to solve...!

Answer 1

Bonjour!

We have to find the volume of acid required to make one liter of 0.1M H2SO4;

Given---›

Percentage of H2SO4 = 98% (w/w)

Density (d) = 1.80 gmL-¹

0.1M of H2SO4

First we will find the normality of the solution:

Formula to find Normality from %(w/w), density and equivalent weight of compound.

$normality = \frac{\%( \frac{w}{w} ) \times d \times 10}{equivalent \: weight \: of \: compound}$

First we will find out equivalent weight of H2SO4 = Weight of compound/Cationic charge on compound

= 98/2

= 49

Therefore normality will be;


$= > normality = \frac{98 \times 1.8 \times 10}{49} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \times 18 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 36N \:$

Therefore,

Normality = 36N


Acid required;

N1V1 = N2V2

=> 36V1 = 0.1 × 2 × 1000. (1L = 1000mL)

=> V1 = 200/36

=> V1 = 5.55 mL

Therefore, the volume of acid required to make one liter of 0.1M H2SO4 is 5.55 mL.

Hope this helps...:)

Answer 2

$\huge\underline\mathfrak\blue{Solution}$