Grams needed to prepare the solution
1 answer:
Answer:
grams of solute = 31.5 g
Explanation:
Given data:
Grams of solute = ?
%m/v = 3%
Volume = 1050 mL
Solution:
grams of solute = 3 g of NH₄Cl / 100 mL × 1050 mL
grams of solute = 0.03 g × 1050
grams of solute = 31.5 g
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If one were to match the ratio of atoms of the elements found in this molecular formula of artificial sweetener it would be :
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Nitrogen - 1 atom
Oxygen - 3 atoms.
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