Question: The molarity of \[C{{l}^{-}}\] in an aqueous solution which was \[\left( w/V \right)2%NaCl,4%Cacl,...

Question

The molarity of $C{{l}^{-}}$ in an aqueous solution which was $\left( w/V \right)2%NaCl,4%Cacl,6%N{{H}_{4}}Cl$ will be
A. $0.342$
B. $0.721$
C. $1.12$
D. $2.18$

Answer 1

Solution

Percent of weight of solution in the absolute volume of solution. Percent here is the number of grams of solute in 100 mL of solution. $W/V$ implies Weight/Volume. This is regularly utilized when a solid is broken down in a liquid. This is used when a solid chemical is dissolved in a liquid.

Complete step by step solution:
Molarity (M) is characterized as the number of moles of solute (n) divided by the volume (V) of the solution in liters. Note that the molarity is characterized as moles of solute per liter of solution, not moles of solute per liter of solvent. Molarity is otherwise called the molar concentration of a solution.
An aqueous solution is one in which the dissolvable is liquid water. That is, solute (disintegrated) ions and molecules are encircled by water molecules and incorporated into the organization of bonds inside the water. The dissolved species at that point spread all through the water. Like a solute of table salt, or sodium chloride $NaCl$ in water it would be represented as $N{{a}^{+}}\left( aq \right)+C{{l}^{-}}\left( aq \right)$
$2%NaCl$ has mole of $C{{l}^{-}}=\dfrac{2}{58.5}$
Same with $4%Cacl$ has mole of $C{{l}^{-}}$$$$=\dfrac{4}{111}$
$6%N{{H}_{4}}Cl$$$$=\dfrac{6}{53.5}$
Total mole in $100ml$$$$=\dfrac{2}{58.5}+\dfrac{4}{111}+=\dfrac{6}{53.5}$
Now, in 1 liter of the solution $2.18mole$
Hence, molarity is $2.18$

So, the correct option is D. $2.18$ .

Note: In aqueous solution, it is highly soluble much of the time; in any case, for some chloride salts, for example, silver chloride, lead (II) chloride, and mercury(I) chloride, they are marginally soluble in water. In aqueous solution chloride is limited by the protic end of the water molecules.