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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` |
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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` |
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