Part A: 80 g of NaOH in 400 g of NaOH solution...needs to be in...

Question

Question

Part A: 80 g of NaOH in 400 g of NaOH solution...needs to be in this format: m/m = ? %

Part B: 3.3 g of KOH and 26.5 g of H2O...………m/m = ? %

Part C: 49.0 g Na2CO3 in 256.5 g of Na2CO3 solution.....m/m = ? %

science chemistry

Answer 1

Answer #1

A)
Given:
Mass of NaOH = 80 g
Mass of solution = 400 g

Use:
(m/m)% = mass of NaOH * 100 / mass of solution
= 80 * 100 / 400
= 20 %
Answer: 20 %

B)
Given:
Mass of KOH = 3.3 g
Mass of solution = 3.3 g + 26.5 g = 29.8 g

Use:
(m/m)% = mass of KOH * 100 / mass of solution
= 3.3 * 100 / 29.8
= 11.1 %
Answer: 11 %

C)
Given:
Mass of Na2CO3 = 49.0 g
Mass of solution = 256.5 g

Use:
(m/m)% = mass of Na2CO3 * 100 / mass of solution
= 49.0 * 100 / 256.5
= 19.1 %
Answer: 19.1 %

Answer 2

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