Calculate the molarity and normality of a solution containing 9.8 g
of H2SO4 in 500 cm3 of the solution.
Answers
Answer:
Molarity of the solution = 0.4 M.
Normality of the solution = 0.8 N.
Step-by-step explanation :
Given that,
Mass of \bold{H_2SO_4}H
2
SO
4
dissolved = 9.8 g.
Volume of the solution = 250 cm³ = 0.250 L.Calculation of molarity :
Molar mass of \bold{H_2SO_4=98\:g\:mol^{-1}}H
2
SO
4
=98gmol
−1
∴ No. of moles of \bold{H_2SO_4=\frac{Mass\:in\:g}{Molar\:mass}}H
2
SO
4
=
Molarmass
Massing
\implies\bold{\dfrac{9.8\:g}{98\:g\:mol^{-1}}=0.1\:mole}⟹
98gmol
−1
9.8g
=0.1mole
Now, \bold{Molarity=\frac{No.\:of\:moles\:of\:the\:solute}{Volume\:of\:solution\:in\:litres}}Molarity=
Volumeofsolutioninlitres
No.ofmolesofthesolute
\implies\bold{\dfrac{0.1\:mol}{0.250\:L}=0.4\:mol\:L^{-1}}⟹
0.250L
0.1mol
=0.4molL
−1
\implies\bold{0.4\:M}⟹0.4MCalculation of normality :
Eq. mass of \bold{H_2SO_4=\frac{Mol.\:mass\:of\:H_2SO_4}{Basicity\:of\:H_2SO_4}}H
2
SO
4
=
BasicityofH
2
SO
4
Mol.massofH
2
SO
4
\implies\bold{\dfrac{98}{2}=49}⟹
2
98
=49
∴ No. of g equivalent of \bold{H_2SO_4=\frac{Mass\:in\:g}{Eq.\:mass}}H
2
SO
4
=
Eq.mass
Massing
\implies\bold{\dfrac{9.8}{49}=0.2}⟹
49
9.8
=0.2
Now, \bold{Normality=\frac{No.\:of\:g\:eq.\:of\:the\:solute}{Volume\:of\:solution\:in\:litres}}Normality=
Volumeofsolutioninlitres
No.ofgeq.
Answer:
To find molarity and normality
Explanation:
As we know that
Molarity = w/molar mass * 1000/volume
Substituting the values
=9.8/98 *1000/500
=0.1 * 10
= 1 m
Normality = M * nf
H2so4 has nf as 2
Therefore
=1 *2
2 N