Answer to Question #127037 in Inorganic Chemistry for Lisakhanya

Question

1. What volume of 68% HNO3 would you need to make up 425 mL of 0.150M HNO3? The density of 68% HNO3 is 1.41 g/mL


2. An experiment requires 0.949g of H2SO4. What volume (in mL) of 0.215M H2SO4 must be used ?

Answer 1

Q.1

Given data:

Volume=425 ml = .425 L

Molarity (M) = .150M

Density of 68% HNO₃= 1.41 g/ml

Molarity = no.of mole/volume of solution in L

No.of mole= Molarity × volume

No.of mol = .150×.425

=.06375 mole

Weight of HNO₃= mole×molecular weight

= .06375×63= 4.01 g

Density= Mass/ volume

Volume required for 4.01 g of 68% HNO3 is

= 4.01/1.41

= 2.85 ml

Q.2

Given data:

Molarity=.215M

Volume=?

Weight of H₂SO₄=.949g

Mole of H₂SO₄= .949/98

=.00968 mole

Molarity= no.of mole/ volume in L

Volume = .00968/.215

Volume= .0450 L

volume in ml

= .0450×1000= 45 ml