A) \[1000\text{ }c{{m}^{3}}\]
B) \[2000\text{ }c{{m}^{3}}\]
C) \[100\text{ }c{{m}^{3}}\]
D) \[200\text{ }c{{m}^{3}}\]
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 neutralisation of \[NaOH=\frac{0.1\times 1000}{1}\] \[=100\text{ }c{{m}^{3}}\]You need to login to perform this action.
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