Question

Molarity of $1m$ aqueous $NaOH$ solution [density of the solution is $1.02gm{{l}^{-1}}$ ]
A) $1M$
B) $1.02M$
C) $1.2M$
D) $0.98M$

Answer 1

In this question, $1m$ is defined as when $1$ moles of solute present in $1kg$ of the solvent then it is known as the $1$ molal solution. The SI unit of molality is $m$ or molal. The commonly used unit of the molality is $molk{{g}^{-1}}$.

Complete answer:
In this question, it is given that the density of the solution is $1.02gm{{l}^{-1}}$ .
The molality is $1m$ that means $1$ moles of $NaOH$ in $1kg$ of the $NaOH$ solution.
The molar mass of the $NaOH$ is $40g$ .
To calculate weight of the solute:
$w=n\times M$
Where, $w$ is the weight of the solute
$n$ is the number of moles of the solute
$M$ is the molar mass of the solute

Now, substituting the values in the above formula we get,
$\Rightarrow w=1\times 40$
$\Rightarrow w=40g$
The weight of the solvent is $1kg\Rightarrow 1000g$

To calculate total weight of the solution
${{W}_{solution}}={{W}_{Solvent}}+{{W}_{solute}}$
Where, ${{W}_{Solution}}$ is the weight of the solution
${{W}_{Solvent}}$ is the weight of the solvent
${{W}_{Solute}}$ is the weight of the solute

Now, substituting the values in the above formula we get,
$\Rightarrow {{W}_{solution}}=1000+40$
$\Rightarrow 1040g$

To calculate density,
$\Rightarrow D=\dfrac{M}{V}$
Where, $D$ is the density of the solution
$M$ is the weight of the solution
$V$ is the volume of the solution

Now substituting the value we get,
$\Rightarrow V=\dfrac{1040}{1.02}$
On further solving we get,
$\Rightarrow V=1019.60ml$

Now we will calculate molarity,
$\Rightarrow M=\dfrac{{{n}_{solute}}}{V\left( l \right)}$
Where, $M$ is the molarity
${{n}_{solute}}$ is the number of moles of the solute
$V$ is the volume of the solution

Now, substituting the values we get,
$\Rightarrow M=\dfrac{1\times 1000}{1019.60}$
$\Rightarrow 0.98M$
The molarity of the solution is $0.98M$

The correct answer is option ‘D’.

In this question, we have calculated the weight of solute present in the solution using mole concept formula. After that, the weight of solvent is added with the weight of solute to calculate the weight of solution. Using the density formula, we have substituted the value of weight of solution and then calculated the volume of the solution. Then we have applied a molarity formula and substituted the value of moles of solute and the volume of solution and calculated the molarity of $NaOH$ solution.

Note: To convert the volume of the solution from millilitres to litre, the volume is divided by $1000$.A molarity is defined as the one litre of solution contains one mole of solute.