How many grams of Na2CO3 are present in 11.8 mL of a solution that is 22.0% Na2CO3 by mass? The density of the solution is 1.10
g/mL.
1 answer:
Answer:
The Answer is 2,86 grs Na2CO3
Explanation:
What we have to do is find the mass of Na2CO3 as a pure component or solute. That's because the 11,8 mL are a solution of Na2CO3. This means, the sum between the solute Na2CO3 and water. To find the grams of Na2CO3 as pure component we create a factor series as is shown in the attached file.
Data:
Density of solution (ρ) = 1,10 grs sln Na2CO3/mL sln Na2CO3
Mass Percentage (%) = 22 grs Na2CO3/100 grs sln Na2CO3
The procedure is explained in the attached file
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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⁺