How many mL of a 4% mass/volume Mg(NO3)2 solution would contain 1.2 grams of magnesium nitrate?
1 answer:
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Answer:
1.44 x 10²⁵ ions of Na⁺
Explanation:
Given parameters:
Mass of NaCl = 1.4kg = 1400g
Unknown:
Number of ions of sodium = ?
Solution:
The compound NaCl in ionic form can be written as;
NaCl → Na⁺ + Cl⁻
In 1 mole of NaCl we have 1 mole of sodium ions
Now, let us find the number of moles in NaCl;
Number of moles =
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Number of moles = = 23.93mol
So;
Since 1 mole of NaCl gives 1 mole of Na⁺
In 23.93 mole of NaCl will give 23.93 mole of Na⁺
1 mole of a substance = 6.02 x 10²³ ions of a substance
23.93 mole of a substance = 6.02 x 10²³ x 23.93
= 1.44 x 10²⁵ ions of Na⁺