Question

To prepare a fertilizer, an engineer dilutes a stock solution of sulfuric acid by adding 23.3 L of 4.50 M acid to enough water to make 500. L. What is the mass of sulfuric acid per milliliter of the diluted solution? Use correct Significant Figures. -

Answer 1

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given

M1 = 4.5 M

V1 = 23.3 L

V2 = 500 L

Use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (4.5*23.3)/500

M2 = 0.2097 M

Now find the mass in 1 mL of solution

volume , V = 1 mL

= 1*10^-3 L

number of mol,

n = Molarity * Volume

= 0.2097*0.001

= 2.097*10^-4 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass of H2SO4,

m = number of mol * molar mass

= 2.097*10^-4 mol * 98.086 g/mol

= 2.06*10^-2 g

Answer: 2.06*10^-2 g