The specific gravity of HCl is \[\mathbf{1}.\mathbf{08}\]. What is its Molarity?

Hint: First of all we are going to find the density of HCl using the formula of specific gravity and after that we will do some unit conversions. After doing the unit conversion we will move forward to calculate our molarity.
Formula to be used:
Specific Gravity = $SG$
Density of substance= ${{p}_{subs\tan ce}}$
Density of reference substance (water) = ${{p}_{water}}$
So, $SG$=$\dfrac{{{p}_{subs\tan ce}}}{{{p}_{water}}}$

Complete answer:
Specific Gravity or can even be referred to as relative gravity is dimensionless that can be defined as the density of a substance divided by the density of the water at a specified point of temperature.
Specific gravity can be obtained by dividing the density of a material by the density of water at ${{4}^{\circ }}C$. For the calculation purpose, the density of the material and that of the water has to be expressed in the same units.
It is very common to use the water’s density at ${{4}^{\circ }}C$ as the point of reference because water at this point has the highest ever density which is\[1000\text{ }kg{{m}^{-3}}\].
To calculate the molarity, we are going to follow the below steps:
Given,
$SG$=$1.08$
${{p}_{water}}$=$1.00g/ml$
So,
$SG$=$\dfrac{{{p}_{HCl}}}{{{p}_{water}}}$
${{p}_{HCl}}$=$SG$$*$${{p}_{water}}$
${{p}_{HCl}}$= $1.08$$*$$1.00g/ml$
${{p}_{HCl}}$=$1.08g/ml$
HCl solution which do have a density of $1.08g/ml$ (room temp is assumed) is near about \[17%w/w\] HCl, that is, you will obtain about \[17\text{ }g\] of hydrochloric acid for \[100.0g\]of the solution.
So, the further steps are as followed for conversion purpose:
\[1.00L*\text{ }\dfrac{1000mL}{1L}*\dfrac{1.08\text{ }g\text{ }}{1ml}=\text{ }1080\text{ }g\] \[1080g\text{ }solution\text{ *}\dfrac{17\text{ }g\text{ }HCl}{100g\text{ }solution}\text{ }=\text{ }183.6\text{ }g\]
\[83.6g*\dfrac{1\text{ }mole\text{ }HCl}{36.46g}=5.0357\text{ }\]moles HCl
So, \[n=5.0357\text{ }\] moles HCl
Now according to the definition of molarity(m), it is the amount of a substance in a specific volume of solution. Molarity can even be stated as the moles of a solute per litre of a specified solution.
So here,
m= $\dfrac{n}{V}$
m= \[\dfrac{5.0357\text{ moles}}{1.00L}\]
So,
m= \[5.0357\text{ M}\]
Therefore, the molarity is \[5.0357\text{ M}\]

Note:
Be very careful with the equations and calculations involved in these types of questions. Here the concept of molarity and specific gravity has been used so give extra attention to the units as well. Also, be ready for the conversion of units in these types of questions.