Question

14. How many mL of a 0.045 M BaCl, contain 15.0 g BaCl2
(208 g mol-')?
(a) 502.45 mL
(b) 1602.6 mL
(c) 726.76 mL
(d) 1241.11 mL
​

Answer 1

Answer:

Explanation:

1 ml=0.045

So

Bacl2=2*0.045

=0.0 90ml

Now

15.0 *280*/0.90

= 1602. 6ml

Hope you like my answer

Answer 2

Answer:

Explanation:

1 ml=0.045

Then,

Bacl2=2×0.045

=0.0 90ml

Therefore,

15.0 ×280×0.90

= 1602. 6ml

Hope it will help you