Question

Calculate the molarity and normality of a solution containing 9.8 g of H2SO4 in 250 cm3 of the solution.​

Answer 1

Answer :

• Molarity of the solution = 0.4 M.
• Normality of the solution = 0.8 N.

Step-by-step explanation :

Given that,

• Mass of $\bold{H_2SO_4}$ dissolved = 9.8 g.
• Volume of the solution = 250 cm³ = 0.250 L.

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✒ Calculation of molarity :

Molar mass of $\bold{H_2SO_4=98\:g\:mol^{-1}}$

∴ No. of moles of $\bold{H_2SO_4=\frac{Mass\:in\:g}{Molar\:mass}}$

$\implies\bold{\dfrac{9.8\:g}{98\:g\:mol^{-1}}=0.1\:mole}$

Now, $\bold{Molarity=\frac{No.\:of\:moles\:of\:the\:solute}{Volume\:of\:solution\:in\:litres}}$

$\implies\bold{\dfrac{0.1\:mol}{0.250\:L}=0.4\:mol\:L^{-1}}$

$\implies\bold{0.4\:M}$

✒ Calculation of normality :

Eq. mass of $\bold{H_2SO_4=\frac{Mol.\:mass\:of\:H_2SO_4}{Basicity\:of\:H_2SO_4}}$

$\implies\bold{\dfrac{98}{2}=49}$

∴ No. of g equivalent of $\bold{H_2SO_4=\frac{Mass\:in\:g}{Eq.\:mass}}$

$\implies\bold{\dfrac{9.8}{49}=0.2}$

Now, $\bold{Normality=\frac{No.\:of\:g\:eq.\:of\:the\:solute}{Volume\:of\:solution\:in\:litres}}$

$\implies\bold{\dfrac{0.2\:g\:eq}{0.250\:L}=0.8\:g\:eq\:L^{-1}}$

$\implies\bold{0.8\:N}$