Question

A student dissolves 15.g of styrene in 250.mL of a solvent with a density of 0.88g/mL. The student notices that the volume of the solvent does not change when the styrene dissolves in it.Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.molarity=______?molality=______?

Answer 1

Answer: The molarity and molality of styrene solution is 0.58 M and 0.66 m respectively

Step-by-step explanation:

To calculate the mass of solvent, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Density of solvent = 0.88 g/mL

Volume of solvent = 250. mL

Putting values in above equation, we get:

`0.88g/mL=\frac{\text{Mass of solvent}}{250mL}\\\\\text{Mass of solvent}=(0.88g/mL* 250mL)=220g`

• Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

Given mass of styrene = 15. g

Molar mass of styrene = 104.15 g/mol

Volume of solution (solvent) = 250. mL

Putting values in above equation, we get:

`\text{Molarity of solution}=(15* 1000)/(104.15* 250)\\\\\text{Molarity of solution}=0.576M=0.58M`

• Calculating the molality of solution:

To calculate the molality of solution, we use the equation:

`Molality=\frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ in grams}}`

Where,

`m_(solute)` = Given mass of solute (styrene) = 15 g

`M_(solute)` = Molar mass of solute (styrene) = 104.15 g/mol

`W_(solvent)` = Mass of solvent = 220 g

Putting values in above equation, we get:

`\text{Molality of styrene solution}=(15* 1000)/(104.15* 220)\\\\\text{Molality of styrene solution}=0.655m=0.66m`

Hence, the molarity and molality of styrene solution is 0.58 M and 0.66 m respectively.