A) 0.2
B) 0.01
C) 0.1
D) 0.3
Correct Answer: C
Solution :
Given \[{{H}_{2}}S{{O}_{4}}-V=100\,\,mL\,\,N=0.2\,M\] \[NaOH-V=100\,\,mL\,\,N=0.2\,M\] Milliequivalent of \[{{H}_{2}}S{{O}_{4}}=100\times 0.2\times 2=40\] (\[\therefore \] it is dibasic acid) Milliequivalent of \[NaOH=100\times 0.1\times 1=20\] \[\therefore \] Milliequivalent of \[{{H}_{2}}S{{O}_{4}}\] left = 40 - 20 = 20 Total volume = 100 mL + 100 mL = 200 mL Normality of \[{{H}_{2}}S{{O}_{4}}\] (left) \[=\frac{20}{200}=0.1\,\,N\]
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