Question

How many grams of $NaHC{O}_3$ are required to neutralise $1mL$ of $0.0902N$ vinegar?

• A. $8.4\times {10}^{-3}g$
• B. $1.5\times {10}^{-3}g$
• C. $0.758\times {10}^{-3}g$
• D. $1.07\times {10}^{-3}g$

Answer 1

Answer: (C)

$0.758\times {10}^{-3}g$

Number of equivalent of $NaHC{O}_3$
$=$ Number of equivalents of acid
$=\frac{NV}{1000}=\frac{0.0902\times 1}{1000}$
Mass of $NaHC{O}_3=\frac{0.0902\times 1\times 84}{1000}$
$=0.758\times {10}^{-3}g$.