Question

the density of stock solution of 49% (w/w) H2SO4 is 1.5 g/ml. the volume of this stock solution required to prepare 2.5L of 0.1 M H2SO4 solution is

Answer 1

Answer:- 33.4 mL

Solution:- Stock solution is 49%(w/w). It means 49 grams of sulfuric acid present in 100 grams of the solution. We want to make 2.5 L of 0.1 M sulfuric acid solution from the stock solution. let's calculate the moles and then grams of this diluted solution from it's given volume and molarity as:

$2.5L(\frac{0.1mol}{1L})(\frac{98g}{1mol})$

= 24.5 g

Density of the stock solution is 1.5 gram per mL. Let's calculate the volume of the stock solution for it's 49% (w)w solution as:

$100g(\frac{1mL}{1.5g})$

= 66.7 mL

It means 49 grams of sulfuric acid are present in 66.7 mL solution. We need to calculate the volume of this stock solution that would contains 24.5 grams of sulfuric acid.

$24.5gH_2SO_4(\frac{66.7mL}{49g})$

= $33.4mLH_2SO_4$

So, 33.4 mL of stock solution are required to make 2.5 L of 0.1 M sulfuric acid solution.

Answer 2

Answer:

Stock solution is 49%(w/w). It means 49 grams of sulfuric acid present in 100 grams of the solution. We want to make 2.5 L of 0.1 M sulfuric acid solution from the stock solution. let's calculate the moles and then grams of this diluted solution from it's given volume and molarity as:

2.5L(\frac{0.1mol}{1L})(\frac{98g}{1mol})2.5L(

1L

0.1mol

)(

1mol

98g

)

= 24.5 g

Density of the stock solution is 1.5 gram per mL. Let's calculate the volume of the stock solution for it's 49% (w)w solution as:

100g(\frac{1mL}{1.5g})100g(

1.5g

1mL

)

= 66.7 mL

It means 49 grams of sulfuric acid are present in 66.7 mL solution. We need to calculate the volume of this stock solution that would contains 24.5 grams of sulfuric acid.

24.5gH_2SO_4(\frac{66.7mL}{49g})24.5gH

2

SO

4

(

49g

66.7mL

)

= 33.4mLH_2SO_433.4mLH

2

SO

4

So, 33.4 mL of stock solution are required to make 2.5 L of 0.1 M sulfuric acid solution.