
68.4g of $A{l_2}{(S{O_4})_3}$ is dissolved in enough water to make $500mL$ of solution. If aluminium sulphate dissociates completely then molar concentration of aluminium ions $A{l^{3 + }}$ and sulphate ions $SO_4^{2 - }$in the solution are respectively (molar mass of $A{l_2}{(S{O_4})_3}$ = $342g$
Answer
557.4k+ views
Hint: Molar concentration of ions will be a product of the molar concentration of the molecule with its stoichiometric coefficient. Molar concentration can be calculated by relating it with the number of moles per unit volume.
Formula used:$n = \dfrac{w}{M}$
Where,
$n$= number of moles
$w$= given mass
$M$= molecular mass
$M = \dfrac{n}{V}$
Where,
$M$= Molarity
$V$= volume
$n$= number of moles
Complete step by step answer:
We know that
$n = \dfrac{w}{M}$
Where,
$n$= number of moles
$w$= given mass
$M$= molecular mass
From the question, we know that
$w$= $68.4g$
$M$= $342g$
Now, substituting these values, we get:
$n = \dfrac{{68.4}}{{342}}$
We get $n$= $0.2$ moles.
Now, we know
$M = \dfrac{n}{V}$
Where,
$M$= Molarity
$V$= volume
$n$= number of moles
From the question we know that:
$V$= $500mL$
Substituting these values in the above equation
$M = \dfrac{{0.2}}{{500mL}}$
This will be equal to:
The molarity will be $0.4M$
Now, let us see the disassociation of the molecule
$A{l_2}{(S{O_4})_3} \to 2A{l^{ + 3}} + 3SO_4^{2 - }$
So the stoichiometric coefficient for $A{l^{3 + }}$ is $2$
So the stoichiometric coefficient for $SO_4^{2 - }$ is $3$
So the molar stoichiometric coefficient for the ions will be
For $A{l^{3 + }}$= $2 \times M$
Where, $M$= Molarity
$2 \times 0.4 \Rightarrow 0.8M$
For $SO_4^{2 - }$= $3 \times M$
Where, $M$= Molarity
$3 \times 0.4 \Rightarrow 1.2M$
Note: Number of moles is the ratio of the given mass to the molecular mass. Avogadro’s number gives a relationship between the number of moles and the number of atoms also. the concept of moles were first introduced because even a very small quantity of an element has an extraordinarily large number of atoms. Hence, in order to interpret data on the basis of this number was rather difficult. Avogadro’s number is a constant value which states every species in this world which has $6.022 \times {10^{23}}$ will make up $1$ mole.
Formula used:$n = \dfrac{w}{M}$
Where,
$n$= number of moles
$w$= given mass
$M$= molecular mass
$M = \dfrac{n}{V}$
Where,
$M$= Molarity
$V$= volume
$n$= number of moles
Complete step by step answer:
We know that
$n = \dfrac{w}{M}$
Where,
$n$= number of moles
$w$= given mass
$M$= molecular mass
From the question, we know that
$w$= $68.4g$
$M$= $342g$
Now, substituting these values, we get:
$n = \dfrac{{68.4}}{{342}}$
We get $n$= $0.2$ moles.
Now, we know
$M = \dfrac{n}{V}$
Where,
$M$= Molarity
$V$= volume
$n$= number of moles
From the question we know that:
$V$= $500mL$
Substituting these values in the above equation
$M = \dfrac{{0.2}}{{500mL}}$
This will be equal to:
The molarity will be $0.4M$
Now, let us see the disassociation of the molecule
$A{l_2}{(S{O_4})_3} \to 2A{l^{ + 3}} + 3SO_4^{2 - }$
So the stoichiometric coefficient for $A{l^{3 + }}$ is $2$
So the stoichiometric coefficient for $SO_4^{2 - }$ is $3$
So the molar stoichiometric coefficient for the ions will be
For $A{l^{3 + }}$= $2 \times M$
Where, $M$= Molarity
$2 \times 0.4 \Rightarrow 0.8M$
For $SO_4^{2 - }$= $3 \times M$
Where, $M$= Molarity
$3 \times 0.4 \Rightarrow 1.2M$
Note: Number of moles is the ratio of the given mass to the molecular mass. Avogadro’s number gives a relationship between the number of moles and the number of atoms also. the concept of moles were first introduced because even a very small quantity of an element has an extraordinarily large number of atoms. Hence, in order to interpret data on the basis of this number was rather difficult. Avogadro’s number is a constant value which states every species in this world which has $6.022 \times {10^{23}}$ will make up $1$ mole.
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