Question

8.
Calculate the quantity of sodium carbonate (anhydrous)
required to prepare 250 mL of 0.1 M solution.
(a) 2.65 g
(b) 4.95 g
(c) 6.25 g
(d) 8.15 g

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Answer 1

$m = \frac{wb}{mb} \times \frac{1000}{vs}$

$0.1 = \frac{wb}{83} \times \frac{1000}{250}$

wb=2.65g

option a

Answer 2

Answer:

see the attachment

thanks ❤️