Question

how many ml of 0.1% w/v solution of mercury bichloride cen be prepare 20 tablet each contain 0.47 g​

Answer 1

Answer:

Step-by-step explanation:

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.