The new concentrations of
and
are 0.25M and 19M
Calculation of number of moles of each component,
Molarity of
= number of moles/volume in lit = 0. 500 M
Number of moles = molarity of
× volume in lit = 0. 500 M× 0.025 L
Number of moles of
= 0.0125 mole
Molarity of
= number of moles/volume in lit = 0. 38 M
Number of moles = molarity of
× volume in lit = 0. 38 M× 0.025 L
Number of moles of
= 0.95 mole
Calculation of new concentration at volume 50 ml ( 0.05L)
Molarity of
= number of moles/volume in lit = 0.0125 mole/0.05L
Molarity of
= 0.25M
Molarity of
= number of moles/volume in lit = 0.95mole/0.05L
Molarity of
= 19 M
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The question here is solved using basic chemistry. CaCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of CaCl2 that dissolves.
CaCl2(s) --> Ca+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.65 mol CaCl2/1L × 2 mol Cl⁻ / 1 mol CaCl2 = 1.3 M
The answer to this question is [Cl⁻] = 1.3 M
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