Question

Sulphuric acid is 95.8% by mass . Calculate the mole fraction and molarity of H2SO4 of density 1.91 cm-3 where H=1,S=32,O=16

Answer 1

For mole Fraction:-
No. of moles of H2SO4=95.8/98=0.9776
No. of Moles of H2O= 4.2/18=0.2333
Mole Fraction=$\frac{moles of H2SO4}{Total moles}$=0.8073

For Molarity:-
Volume of acid=100/1.91=52.3560 cm^3
molarity= $\frac{95.8}{98}* \frac{1000}{52.3560}$=18.6712

Answer 2

Hey dear,

● Answer -
X2 = 0.8073
M = 18.10 molal

● Explanation -
Sulphuric acid soln is 95.8 % W/W.
Thus, in 100 g H2SO4 sol -
Water => W1 = 4.2 g
H2SO4 => W2 = 95.8 g


No of moles are calculated by -
Water => n1 = 4.2/18 = 0.2333
H2SO4 => n2 = 95.8/98 = 0.9776

Mole fraction of solute is calculated by -
X2 = 0.9776 / (0.2333 + 0.9776)
X2 = 0.8073

Molality of H2SO4 solution is given by -
M = n2 / V
M = 0.9776 / 0.052
M = 18.10 molal

Hope this helps you...