Question

A teacher prepared HCl solution by dissolving 1.46 g of
HCl is 250 ml of water and Na_CO3 solution by
dissolving 2.12 g of Na2CO3 in 1L water. Then she
asked her students to perform the titration. The
amount of HCl that would be required to neutralise 20
ml of Na2CO3 solution will be
0.08 g
2.989
0.008 g
0.03 g​

Answer 1

Answer:

2.98g

Explanation:

molarity of HCl solution

= 1.46g/36.5g/mol /0.250L

= 0.16M

molarity of Na2CO3 solution = 2.12g/106g/mol

= 0.02M

let V ml of HCl required.

n - factor×M1V1 = M2V2× n-factor

1× 0.16M×v = 0.02M × 20 ml ×2

V = 5 ml

wt(amount) of HCl required = 2.9 g

Answer 2

Answer:

HCI = 2.9 g.................