Question

# What is the Molarity of $HCl$ in a solution prepared by dissolving $5.5{\text{ }}g{\text{ }}HCl$ in $200g$ of ethanol? If the density of solution is$0.79g/ml$ ?A. $0.58M$ B. $0.21M$ C. $0.93M$ D. $1.7M$

Hint:Molarity is defined as the number of moles of solute dissolved in one litre of the solution i.e. molarity is equal to the ratio of the weight of solute to the product of molecular weight of solute and volume of solution.
Formula used: Molarity $= \dfrac{{number\,of\,moles\,of\,solute}}{{{\text{Volume of solution}}}} \times 1000$

Given: Weight of solute ${\text{(HCl) = 5}}{\text{.5 g}}$
Weight of solvent (ethanol) ${\text{ = 200 g}}$
Density of solution ${\text{ = 0}}{\text{.79 g/ml}}$
Mass of solution $=$ Mass of $HCl\,\, +$Mass of ethanol
$= {\text{ }}5.5 + 200 \\ = {\text{ }}205.5{\text{ }}gm{\text{.}} \\$
We know that molarity is defined as the number of moles of solute dissolved per liter of the solution.
i.e. Molarity $= \,\dfrac{{number\,of\,moles\,of\,solute}}{{{\text{Volume of solution (ml)}}}} \times 1000$
$\therefore$ Molarity $= \dfrac{{{\text{Weight of solute}}}}{{Molecular{\text{ weight of }}\,solute \times \,Volume\,of\,solution\,(ml)}} \times 1000$ …………. (i)
($\because$ Number of moles $= \,\dfrac{{weight\,of\,solute}}{{Molecular\,weight\,of\,solute}}$ )
Molarity$= {\text{ }}\dfrac{{{\text{Weight of HCl }} \times {\text{ Density of solution}}}}{{Molecular{\text{ weight of HCl }} \times {\text{ Mass of solution}}}} \times 1000$…………………. (ii)
($\because$ Volume of solution${\text{ = }}\dfrac{{Mass{\text{ of solution}}}}{{Density{\text{ of solution}}}}$)
In equation (ii), putting the value of weight of $HCl,$density of solution, molecular weight of $HCl,$and mass of solution, then we get,
Molarity $= \dfrac{{5.5 \times 0.79 \times 1000}}{{36.5 \times 205.5}}$ [$\because$ Molecular weight of $HCl$ is $36.5gm$]
On solving the equation
Thus molarity $= \,0.579M$

Hence the correct answer is option (A).

Note:Molarity of a solution depends upon temperature because volume of a solution is temperature dependent. When a concentrated solution is diluted by adding more solvent, the number of moles of solute in the solution remains unchanged.