A) \[1000\,\,c{{m}^{3}}\]
B) \[2000\,\,c{{m}^{3}}\]
C) \[100\,\,c{{m}^{3}}\]
D) \[200\,\,c{{m}^{3}}\]
Correct Answer: C
Solution :
Weight of pure\[NaCl=6.5\times 0.9=5.85\,\,g\] No. of equivalence of\[NaCl=\frac{5.85}{58.5}=0.1\] No. of equivalence of\[NaOH\]obtained\[=0.1\] Volume of 1 M acetic acid required for the neutralization of\[NaOH=\frac{0.1\times 1000}{1}\] \[=100\,\,c{{m}^{3}}\]You need to login to perform this action.
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