What mass of sodium carbonate is required to prepare 0.5 liters of a 13% solution with a density of 1.13 g / ml?
| January 29, 2021 | education
1. Let’s find the mass of the solution we need:
0.5 l = 500 ml.
m (solution) = V (solution) * ρ (solution) = 500 ml * 1.13 g / ml = 565 g.
2. Find the mass of sodium carbonate in solution:
m (Na2CO3) = m (solution) * ω (Na2CO3) / 100% = 565 g * 13% / 100% = 73.45 g.
Answer: m (Na2CO3) = 73 g.
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