Question

Calculate the molarity of each of the ions in a solution when 3.0 litre of 4.0 M \[NaCl\] and 4.0 litre of 2.0 M \[CaC{l_2}\] are mixed and diluted to 10 litre.

Answer 1

Hint: Molarity of a given solution is defined as the total number of moles of solute per litre of solution.
\[M = \dfrac{n}{V}\]
Here,
\[M\] is the molality of the solution that is to be calculated
\[n\] is the number of moles of the solute
\[V\] is the volume of solution given in terms of litres

Complete step by step answer:
To proceed with the question let us 1st find the number of moles of moles of \[N{a^ + }\], \[C{a^{2 + }}\] and \[C{l^ - }\] present in the given solution,
Therefore,
Number of moles of \[N{a^ + }\] ions = Molarity $ \times $ Given Volume = \[3 \times 4{\text{ }} = 12\]
Number of moles of \[C{a^{2 + }}\] ions = Molarity $ \times $ Given Volume = \[2 \times 4{\text{ }} = 8\]
Number of moles of \[C{l^ - }\] ions = Molarity $ \times $ Given Volume = \[3 \times 4 + 2 \times 4 \times 2{\text{ }} = 28\](because \[C{l^ - }\] ions are present in both solution, therefore we need to add the number of moles of \[C{l^ - }\] of both solution in order to find the number of moles).
As per the given question total volume of the mixture = 10L
Therefore, molarity of \[N{a^ + }\], \[C{a^{2 + }}\] and \[C{l^ - }\] in 10L of solution is given by the formula =
\[M = \dfrac{n}{V}\]
Where,
\[M\] is the molality of the solution that is to be calculated
\[n\] is the number of moles of the solute
\[V\] is the volume of solution given in terms of litres
Molarity of \[N{a^ + }\] ions =\[M = \dfrac{n}{V} = \dfrac{{12}}{{10}} = 1.2{\text{M}}\]
Molarity of \[C{a^{2 + }}\] ions =\[M = \dfrac{n}{V} = \dfrac{8}{{10}} = 0.8{\text{M}}\]
Molarity \[C{l^ - }\] ions =\[M = \dfrac{n}{V} = \dfrac{{28}}{{10}} = 2.8{\text{M}}\]
Note:
Molarity is the number of moles of a solute per litre of solution and Molality is the number of moles of solute in 1 kg of solvent. Therefore, the relationship between Molarity and Molality is given by:
$m = \dfrac{{M \times 1000}}{{(d \times 1000) - M \times {M'}}}$
Where,
$m$= Molality
$M$= Molarity
$d$= Density
${M'}$= Molar mass of solute