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Question: Two solutions of \(1M\) \(CaC{l_2}\) and \(2M\) \(AlC{l_3}\) were mixed in volume ratio of \(1:2\) ,...

Two solutions of 1M1M CaCl2CaC{l_2} and 2M2M AlCl3AlC{l_3} were mixed in volume ratio of 1:21:2 , calculate the concentration of ClC{l^ - } in the resulting mixture:
A. 4.20M4.20M
B. 4.50M4.50M
C. 4.66M4.66M
D. 4.95M4.95M

Explanation

Solution

Calcium chloride (CaCl2CaC{l_2}) is a chemical compound consisting of one calcium divalent ion and two chloride monovalent ions in it. Similarly, aluminium chloride (AlCl3AlC{l_3}) is a chemical compound consisting of one aluminium trivalent ion and three chloride monovalent ions.

Complete step by step answer:
The dissociation of calcium chloride in the solution into its corresponding ions takes place according to the following reaction:
CaCl2Ca2++2ClCaC{l_2} \to C{a^{2 + }} + 2C{l^ - }
Similarly, the dissociation of aluminium chloride in the solution into its corresponding ions takes place according to the following reaction:
AlCl3Al3++3ClAlC{l_3} \to A{l^{3 + }} + 3C{l^ - }
According to the question, 1M1M CaCl2CaC{l_2} and 2M2M AlCl3AlC{l_3} were mixed in volume ratio of 1:21:2.
Let the volume of calcium chloride be xx litres and the volume of aluminium chloride be 2x2x litres.
As we can see from the dissociation of calcium chloride, 1mole1mole of CaCl2CaC{l_2} gives 2moles2moles of ClC{l^ - } ions.
So, in xx litres of 1M1M of CaCl2CaC{l_2}solution = 2x2x moles of ClC{l^ - } ions are present.
As we can see from the dissociation of aluminium chloride, 1mole1mole of AlCl3AlC{l_3} gives 3moles3moles of ClC{l^ - } ions.
In 2M2M AlCl3AlC{l_3} solution = 2×3=6moles2 \times 3 = 6moles of ClC{l^ - } ions are present.
So, in 2x2x litres of 2M2M of AlCl3AlC{l_3}solution = 2x×6=12x2x \times 6 = 12x litres of ClC{l^ - } ions are present.
Total moles of chloride ions in the resultant mixture of calcium chloride and aluminium chloride = (12x+2x)moles(12x + 2x)moles of ClC{l^ - } ions.
Total volume of mixture = (x+2x)l=3x(x + 2x)l = 3x litres
The concentration of chloride ions = nV\dfrac{n}{V}
Where, n=n = number of moles of chloride ions in the mixture
V=V = volume of mixture
[Cl]=nV=2x+12x3x=143=4.66mol/l[C{l^ - }] = \dfrac{n}{V} = \dfrac{{2x + 12x}}{{3x}} = \dfrac{{14}}{3} = 4.66mol/l
Thus, the correct option is C. 4.66M4.66M .

Note:
In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical description can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration.