Answer : The concentration of solution is, 8.53 M.
Explanation :
As we are given, 45.0 mass % solution of ethanol in water that means 45.0 g of ethanol present in 100 g of solution.
First we have to calculate the volume of solution.
[tex]\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}[/tex]
[tex]\text{Volume of solution}=\frac{100g}{0.873g/mL}=114.5mL[/tex]
Now we have to calculate the molarity of solution.
Mass of [tex]C_2H_5OH[/tex] = 45.0 g
Volume of solution = 114.5 mL
Molar mass of [tex]C_2H_5OH[/tex] = 46.07 g/mole
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Mass of }C_2H_5OH\times 1000}{\text{Molar mass of }C_2H_5OH\times \text{Volume of solution (in mL)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{45.0g\times 1000}{46.07g/mole\times 114.5mL}=8.53mole/L=8.53M[/tex]
Therefore, the concentration of solution is, 8.53 M.