
The molar concentration of chloride ions in the resulting solution of 300ml of 3.0 M NaCl and 200 mL of $4.0 \ M \ BaC{{l}_ {2}} $ will be:
(A) 1.7 M
(B) 1.8 M
(C) 5.0 M
(D) 3.5 M
Answer
571.2k+ views
Hint: Molar concentration is a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution. To calculate the Molar Concentration, we will find the molar concentration by dividing the moles by litres of water used in the solution.
Complete step by step answer:
We have been provided with:
$NaCl \rightleftharpoons Na^+ + Cl^-$
Salts that are soluble in water dissociate into their ions and are electrolytes. Sodium chloride, NaCl, is a water-soluble salt that dissociates totally in water. The process by which this takes place involves the surrounding of each positive sodium ion and each negative chloride ion by water molecules
Where, concentration of NaCl and chloride ions are 3.0 M.
\[BaC{{l}_ {2}} \rightleftharpoons B{{a}^ {2+}} +2C{{l}^ {-}} \]
$BaC{{l}_ {2}} $ in water is both hygroscopic and water-soluble. When exposed to an open flame, the compound gives a yellow-green coloration. The salt is produced by reacting hydrochloric acid with either barium carbonate or barium hydroxide. The result is hydrated $BaC{{l}_ {2}} $.
Where, concentration of $BaC{{l}_ {2}} $ is 4.0 M and that of chloride ion is 8.0 M.
Now, we will be finding the molar concentration using the formula:
Molar concentration is a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.
\[C{{l}^ {-}} =\dfrac{{{ M}_ {1}} {{V}_ {1}} +{{ M}_ {2}} {{V}_ {2}}} {{{V}_ {1}} +{{V}_ {2}}} \]
Now, keeping the values in this formula:
\[C{{l}^ {-}} =\dfrac {3\times 300+2\times 4\times 200} {300+200} \]
Simplifying the value: $C{{l}^ {-}} =\dfrac {900+1600}{500} $ .
So, the molar concentration comes out to be: 5 M
Therefore, we can conclude that option (C) is correct.
Note: The chloride ion is formed when the element chlorine gains an electron or when a compound such as hydrogen chloride is dissolved in water or other polar solvents. Chloride salts such as sodium chloride are often very soluble in water.
Complete step by step answer:
We have been provided with:
$NaCl \rightleftharpoons Na^+ + Cl^-$
Salts that are soluble in water dissociate into their ions and are electrolytes. Sodium chloride, NaCl, is a water-soluble salt that dissociates totally in water. The process by which this takes place involves the surrounding of each positive sodium ion and each negative chloride ion by water molecules
Where, concentration of NaCl and chloride ions are 3.0 M.
\[BaC{{l}_ {2}} \rightleftharpoons B{{a}^ {2+}} +2C{{l}^ {-}} \]
$BaC{{l}_ {2}} $ in water is both hygroscopic and water-soluble. When exposed to an open flame, the compound gives a yellow-green coloration. The salt is produced by reacting hydrochloric acid with either barium carbonate or barium hydroxide. The result is hydrated $BaC{{l}_ {2}} $.
Where, concentration of $BaC{{l}_ {2}} $ is 4.0 M and that of chloride ion is 8.0 M.
Now, we will be finding the molar concentration using the formula:
Molar concentration is a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.
\[C{{l}^ {-}} =\dfrac{{{ M}_ {1}} {{V}_ {1}} +{{ M}_ {2}} {{V}_ {2}}} {{{V}_ {1}} +{{V}_ {2}}} \]
Now, keeping the values in this formula:
\[C{{l}^ {-}} =\dfrac {3\times 300+2\times 4\times 200} {300+200} \]
Simplifying the value: $C{{l}^ {-}} =\dfrac {900+1600}{500} $ .
So, the molar concentration comes out to be: 5 M
Therefore, we can conclude that option (C) is correct.
Note: The chloride ion is formed when the element chlorine gains an electron or when a compound such as hydrogen chloride is dissolved in water or other polar solvents. Chloride salts such as sodium chloride are often very soluble in water.
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