
How would you calculate the molarity of the following solution: $0.900mol$ of $N{{a}_{2}}S$ in $1.10L$ of solution?
Answer
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Hint: The concentration of a solution can be measured by different forms like, molarity, molality, mass by mass percentage of solute etc.
The molarity and molality of a solution, both takes in account the number of moles of the solute, but the molarity establishes a relationship between volume of the solution and number of moles, whereas molality takes in account the amount of solvent.
Formula used:
$M=\dfrac{n}{V}$
Where $M$ is the molarity of the solution, $n$ signifies the number of moles of solute, and $V$, signifies the volume of the solution in terms of litre.
Complete step-by-step answer:At first, we will define the molarity of a solution. It is a measure of concentration of the solute present per litre of the solution. The concentration of the solute is taken, in terms of moles. And so molarity establishes a relationship between the number of moles of solute and the volume of solvent.
Now, consider the question has given some of the values of quantities of number of moles and volume of the solution, of which we have to calculate the molarity. Now, we will write all the values of the given quantities, which are \[n=0.900mol\] which is the number of moles of the solute sodium sulphide, and the volume of the solution is $V=1.10L$. The equation of molarity, as per the definition, can be expressed as,
$M=\dfrac{n}{V}$
Where $M$ is the molarity of the solution, $n$ signifies the number of moles of solute which is sodium sulphide in this case and $V$, signifies the volume of the solution in terms of litre.
Now, we will substitute the values in the equation, in order to get the molarity of the solution,
$M=\dfrac{0.900mol}{1.10L}$. Now, we will solve the equation in order to get the value of molarity which is, $M=0.81mol{{L}^{-1}}$ which is the desired answer.
Note: The molarity of a solution can be calculated by using the number of moles of solute which is present in the given solution, and the volume of the solution. The very definition of the molarity establishes a relationship between both these quantities and so it can be used to determine the molarity.
The molarity and molality of a solution, both takes in account the number of moles of the solute, but the molarity establishes a relationship between volume of the solution and number of moles, whereas molality takes in account the amount of solvent.
Formula used:
$M=\dfrac{n}{V}$
Where $M$ is the molarity of the solution, $n$ signifies the number of moles of solute, and $V$, signifies the volume of the solution in terms of litre.
Complete step-by-step answer:At first, we will define the molarity of a solution. It is a measure of concentration of the solute present per litre of the solution. The concentration of the solute is taken, in terms of moles. And so molarity establishes a relationship between the number of moles of solute and the volume of solvent.
Now, consider the question has given some of the values of quantities of number of moles and volume of the solution, of which we have to calculate the molarity. Now, we will write all the values of the given quantities, which are \[n=0.900mol\] which is the number of moles of the solute sodium sulphide, and the volume of the solution is $V=1.10L$. The equation of molarity, as per the definition, can be expressed as,
$M=\dfrac{n}{V}$
Where $M$ is the molarity of the solution, $n$ signifies the number of moles of solute which is sodium sulphide in this case and $V$, signifies the volume of the solution in terms of litre.
Now, we will substitute the values in the equation, in order to get the molarity of the solution,
$M=\dfrac{0.900mol}{1.10L}$. Now, we will solve the equation in order to get the value of molarity which is, $M=0.81mol{{L}^{-1}}$ which is the desired answer.
Note: The molarity of a solution can be calculated by using the number of moles of solute which is present in the given solution, and the volume of the solution. The very definition of the molarity establishes a relationship between both these quantities and so it can be used to determine the molarity.
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