Question

How many grams of sodium bicarbonate are required to neutralise 10 ml of 0.92 M

vinegar ?

(a) 8.4 g

(b) 1.5 g

(c) 0.758 g

(d) 1.07 g​

Answer 1

Answer:

75 grams. is the right answer of this question

Answer 2

Answer:

Correct option is (c) 0.75g

Explanation:

CH3COOH+NaHCO3→CH3COONa+H2O+CO2

Equivalents of CH3COOH=Molarity×volume×nf=0.902×10×10−3

Equivalents of NaHCO3 required =Moles×nf=0.902×10×10−3

Moles of NaHCO3=9.02×10−3

Weight of NaHCO3=9.02×10−3×84=0.75 grams