Question

How do you determine the percent by molarity and molality of 230 grams of sodium chloride dissolved in 1025.0 g of water? Density of solution is 1.112 g/mL?

Answer 1

Hint: Concentration of a solution can be expressed in terms of molarity and molality. Molarity deals with volume of solution, while molality deals with mass of solvent.
Formula used: For molarity = $\dfrac{moles\,of\,solute}{volume\,of\,solution}$
For molality = $\dfrac{moles\,of\,solute}{ki\log rams\,of\,solvent}$

Complete answer:
We have to find out the molarity and molality of a solution. Given that,
Mass of solute= 230 g
Mass of solution = 1025.0 g
Density of solution = 1.112 g/mL
First we will find molarity, for this volume is needed which has to be determined from the given density as,
\[density=\dfrac{mass}{volume}\]
So, volume = $\dfrac{mass}{density}$ = $\dfrac{230\,g\,+1025\,g}{1.112\,g/m{{L}^{-1}}}$
Volume = 1128.6 mL
Now, using the formula of molarity = $\dfrac{moles\,of\,solute}{volume\,of\,solution}$, we have to take out the number of moles of solute, as we have given its mass.
So, moles=$\dfrac{given\,mass}{molar\,mass}$
Moles of solute = $\dfrac{230\,g}{58.44g\,mo{{l}^{-1}}}$
Moles of solute= 3.93 moles
Keeping them in the molarity formula,
Molarity = $\dfrac{3.93\,moles}{1128.6mL\times {{10}^{-3}}L.m{{L}^{-1}}}$
Molarity = 3.49 mol ${{L}^{-1}}$
Now we have to calculate the molality of this solution, using:
Molality = $\dfrac{moles\,of\,solute}{ki\log rams\,of\,solvent}$
Molality = $\dfrac{3.93\,moles}{1025.0g\times {{10}^{-3}}g.\,K{{g}^{-1}}}$
Molality = 3.84 mol $K{{g}^{-1}}$
Hence, the molarity of the solution is 3.49 mol ${{L}^{-1}}$ and the molality is 3.84 mol $K{{g}^{-1}}$.

Note:
As molarity is calculated in liters, hence in the calculations, volume needs to be multiplied by the conversion faction of ${{10}^{-3}}$. Also the molality needs to be calculated in kilograms, hence in the calculation, mass needs to be multiplied by the factor of ${{10}^{-3}}$.