Question

Concentrated sulfuric acid is 96.0% (w/w) sulfuric acid (H2SO4, MW = 98.1g/mol) and has a density of 1.84 g/mL. Determine the volume needed to prepare 500 mL of a 1.00 M solution.

Answer 1

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 1.84 g/mL

we have below equation to be used:

mass = density * volume

= 1.84 g/mL *1*10^3 mL

= 1840.0 g

This is mass of solution

mass of H2SO4 = 96.0 % of mass of solution

= 96.0*1840.0/100

= 1766.4 g

Molar mass of H2SO4 = 98.1 g/mol

mass of H2SO4 = 1766.4 g

we have below equation to be used:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(1766.4 g)/(98.1 g/mol)

= 18.01 mol

volume , V = 1 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 18.01/1

= 18.01 M

This is initial concentration of H2SO4

use dilution formula

M1*V1 = M2*V2

Here:

M1 is molarity of solution before dilution

M2 is molarity of solution after dilution

V1 is volume of solution before dilution

V2 is volume of solution after dilution

we have:

M1 = 18.01 M

M2 = 1.0 M

V2 = 500.0 mL

we have below equation to be used:

M1*V1 = M2*V2

V1 = (M2 * V2) / M1

V1 = (1*500)/18.01

V1 = 27.8 mL

Answer: 27.8 mL