A) 0.56 N
B) 0.50 N
C) 0.40 N
D) 0.35 N
Correct Answer: A
Solution :
For\[HCl\] For\[{{H}_{2}}S{{O}_{4}}\] \[{{V}_{1}}=20mL\] \[{{V}_{2}}=10\,mL\] \[{{N}_{1}}=10\text{ }N\] \[{{N}_{2}}=36\text{ }N\] \[\therefore \] Normality of resulting mixture \[(N)=\frac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\] \[=\frac{10N\times 20+36N\times 10}{20+10}\] \[=\frac{200N+360N}{30}\] \[=\frac{560N}{30}=18.67\,N\] \[\therefore \]The mixture is made one litre, therefore the normality of this solution is \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[\therefore \] \[18.67\times 30={{N}_{2}}\times 1000\] \[{{N}_{2}}\frac{18.67\times 30}{1000}=0.56\,N\]
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