Question

How many grams of concentrated sulphuric acid
solution [80% (w/w)] should be used to prepare
500 mi of 1 MH,SO?​

Answer 1

The required number of moles=( 250/1000)×2

=0.5 moles

(no. of moles=molarity × volume)

so,required mass of HNO3=0.5×63

=31.5 grams

(given mass=no. of moles × molar mass)

Given,

70 grams of HNO3 are present in 100 grams of the solution

so,1 gram will be present in 100/70 grams of solution

hence 31.5 grams will be present in (100/70)×31.5 grams of solution.

so amount of concentrated nitric acid solution used is 45 grams.

Answer 2

Answer:

The required number of moles=( 250/1000)×2

=0.5 moles

(no. of moles=molarity × volume)

so,required mass of HNO3=0.5×63

=31.5 grams

(given mass=no. of moles × molar mass)

Given,

70 grams of HNO3 are present in 100 grams of the solution

so,1 gram will be present in 100/70 grams of solution

hence 31.5 grams will be present in (100/70)×31.5 grams of solution.

so amount of concentrated nitric acid solution used is 45 grams.