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Answer:
0.52 g of KNO₃ are contained in 19.7 mL of diluted solution.
Explanation:
We can work on this problem in Molarity cause it is more easy.
Molarity (mol/L) → moles of solute in 1L of solution.
100 mL of solution = 0.1 L
We determine moles of solute: 44.7 g . 1mol /101.1 g = 0.442 mol of KNO₃
Our main solution is 0.442 mol /0.1L = 4.42 M
We dilute: 4.42 M . (11.9mL / 200mL) = 0.263 M
That's concentration for the diluted solution.
M can be also read as mmol/mmL, so let's find out the mmoles
0.263 M . 19.7mL = 5.18 mmol
We convert the mmol to mg → 5.18 mmol . 101.1 mg / mmol = 523.7 mg
Let's convert mg to g → 523.7 mg . 1 g / 1000 mg = 0.52 g