Question

The normality of the solution obtained by mixing 250mL of 0.25 M H2SO4 and 250mL of 0.1N HCl is
a) 0.175N
b)0.255N
c)0.3N
d)0.4N​

Answer 1

$n = m \div 2 \\ n = 0.25 \div 2 \\ n = 0.155 \\ n = 0.225$

Answer 2

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

• Molarity of 250ml of $\sf H_2SO_4$ is 0.25 M.
• Normality of 250ml of HCl is 0.1N

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

• The final normality of the solution.

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } FormulA\:UseD:-}}}}}$

We can find Normality using these two formulae:

$\underline{\boxed{\green{\bf \dag \:\: Normality\:=\:n-factor\:\times\: Molarity}}}$

$\tt And \:other\:one\:is:-$

$\underline{\boxed{\green{\bf \dag \:\: N_1V_1+N_2V_2\:=\:N_3(V_1+V_2)}}}$

$\bf Where$

• $\sf N_1\:is\: Normality\:of\:first\:sol^n$
• $\sf N_2\:is\: Normality\:of\:second\:sol^n$
• $\sf V_1\:is\: Volume\:of\:first\:sol^n$
• $\sf V_2\:is\: Volume\:of\:second\:sol^n$
• $\sf N_3\:is\: Normality\:of\:final\:sol^n$

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

Now here given that Molarity of first solⁿ is 0.25 M and it's volume is 250ml .

Firstly we know that the n - factor of Sulphuric acid is 2 . So Normality would be :

$\tt:\implies Normality =n-factor\:\times Molarity$

$\tt:\implies Normality = 2\times 0.</p><p>25$

$\tt:\implies Normality =\cancel{2}\times\dfrac{\cancel{25}}{\cancel{100}}$

$\underline{\boxed{\red{\tt{\longmapsto\:\: Normality\:\:=\:\:\dfrac{1}{2}N}}}}$

________________________________

Now , given that Normality of second solution is 0.1 N . Now , put the respective values in above formula stated ,

$\tt:\implies N_1V_1+N_2V_2=N_3(V_1+V_2)$

$\tt:\implies \dfrac{1}{2}\times\bigg\lgroup \dfrac{\cancel{250}}{\cancel{1000}}\bigg\rgroup+\dfrac{1}{10}\times\bigg\lgroup \dfrac{\cancel{250}}{\cancel{1000}}\bigg\rgroup =N_3\bigg\lgroup\dfrac{\cancel{250}}{\cancel{1000}}+\dfrac{\cancel{250}}{\cancel{1000}}\bigg\rgroup$

$\tt:\implies \dfrac{1}{4}\bigg\lgroup \dfrac{1}{2}+\dfrac{1}{10}\bigg\rgroup =N_3\times\dfrac{1}{2}$

$\tt:\implies \dfrac{\cancel2\times1}{\cancel4}\bigg\lgroup \dfrac{5+1}{10}\bigg\rgroup=N_3$

$\tt:\implies N_3=\dfrac{1}{\cancel2}\times\dfrac{\cancel6}{10}$

$\tt:\implies N_3=\dfrac{3}{10}N$

$\underline{\boxed{\red{\tt{\longmapsto\:\: N_3\:\:=\:\:0.3N}}}}$

$\boxed{\green{\bf\blue{\dag}\:Hence\:the\:final\: Normality\:is\:0.3N.}}$