Question

how would you prepare 1.15 liter of 1.00M HNO3 from a concentrated solution that is 70.3% HNO3 by mass and has a density of 1.41 g/ml?

Answer 1

you need to prepare 1.15 L of 1.00 M HNO3
so,
moles of HNO3 required = M*V
= 1.00 M * 1.15 L
= 1.15 mol

mass of HNO3 required = number of mol * molar mass
= 1.15 mol * 63 g/mol
= 72.45 g

Let say we used v mL of 70.3% HNO3 solution
since density is 1.41 g/mL
mass of solution = 1.41*v g

mass of HNO3 in this solution = 70.3 % of 1.41*v
= 0.703*1.41*v

this mass should equal 72.45 g

0.703*1.41*v = 72.45
v = 73.1 mL

So to prepare the required solution, we need to take 73.1 mL of original solution and dilute it to 1.15 L