Question

The concentration of 9.8 g of H2SO4 in 500 ml solution is ?

(A) 0.5 M (B) 0.2 M (C) 0.4 N (D) 0.02 M

please tell me the formula to find it also?

Answer 1

Normality, N, = M / feq
where,
M is the molar concentration in mol/L
feq is the equivalence factor in Eq
mEq = m, mass of the molecule in mg/(Atomic weight/valence)

Solve for Normality:
H2SO4 = 98g/mol
9.8g H2SO4 * 1mol/98g = 0.1mol H2SO4

Determine Molarity of H2SO4
M = m/V
V = 500mL = 0.500L
m = 0.1mol
M = 0.10mol/0.500L = 0.2mol/L or 0.2M ---- Molarity of 9.8g H2SO4 in 500mL solution

Determine the Equivalence factor
H2 = 2g/mol * 0.1mol H2SO4 = 0.2g = 200mg H atom
S = 32g/mol * 0.1mol H2SO4 = 3.2g = 3200mg S atom
O4 = 64g/mol * 0.1mol H2SO4 = 6.4 = 6400mg O atoms
valences of atoms:
H = 1
S = 2
O = 2
mEq = m/(AW/valence number)
H2 = mEq = 200mg/(2g/1vn) = 100mEq H2
S = mEq = 3200mg/(32/2) = 200mEq S
O4 = mEq = 6400mg/(64/2) = 200mEq
Sum Total = 500mEq H2SO4 * 1Eq/1000mEq = 0.500Eq

Finally, Solve for Normality
N = Ci/feq
N = 0.20mol/L / 0.500Eq = 0.40mol/L or 0.40Eq/L or 400mEq/L or 0.40N