Question

Density of a 3.75 M sulfuric acid (H2SO4) solution is 1.23 g/ml. Calculate its mass %, XH2SO4, molality & normality.

Answer 1

Let's consider 1L of H2SO4.

So volume, V = 1L

and Concentration, M = 3.75 M

=> Moles of H2SO4 present in 1L = M*V = 3.75 mol/L * 1L = 3.75 mol

Molar mass of H2SO4 = 98.08 g/mol

=> mass of H2SO4 in 1L solution = mol * Molar mass = 3.75 mol * 98.08 g/mol = 367.8 g

Given density of solution = 1.23 g/mL

=> Mass of 1L(=1000mL) H2SO4 solution = Volume * Density = 1000 mL * 1.23 g/mL = 1230 g

(a): Hence mass % = $\frac{mass of H2SO4}{Mass of solution}*100$

= $\frac{367.8 g}{1230 g}*100$

= 29.9 % (answer)

(b): Moles of H2SO4 in 1L = 3.75 mol

Mass of water in 1L solution = Mass of solution - Mass of H2SO4

= 1230 g - 367.8 g

= 862.2 g H2O

=> Moles of water = mass / molar mass = 862.2 g / 18.015 g/mol = 47.86 mol

Total moles = Moles of H2SO4 + Moles of H2O = 3.75 mol + 47.86 mol = 51.61 mol

=> Mole fraction of H2SO4, X_H2SO4 $=\frac{Moles of H2SO4}{Total moles}$

$=\frac{3.75 mol}{51.61 mol}$   

= 0.0727 (answer)

(c): The formula for molality is

$Molality, m = \frac{Moles of H2SO4}{grams of H2O}*1000$

$=\frac{3.75}{862.2}*1000$

= 4.35 molal (answer)

(d): Since H2SO4 has 2 replacible H, n-factor = 2

=> Normality = Molarity * n-factor = 3.75 * 2 = 7.5 N (answer)