1.

How many grams of `NaHCO_(3)` are required to neutralise 1 mL of 0.0902 N vinegar ?A. `8.4 xx 10^(-3)g`B. `1.5xx10^(-3)g`C. `0.758xx10^(-3)g`D. `1.07xx10^(-3)g`

Answer» Correct Answer - C
Number of equivalents of `NaHCO_(3)`
=Number of equivalents of acid
`=(NV)/(1000)=(0.09202xx1)/(1000)`
Mass of `NaHCO_(3)=(0.09202xx1xx84)/(1000)`
`=0.758xx10^(-3)g`


Discussion

No Comment Found

Related InterviewSolutions