Question

A solution is prepared by dissolving 42.0 g of glycerin, C3H8O3, in 186 g of water with a final volume of 200.0 mL. a. Calculate the molarity and molality of the solution. b. What would be the molarity if 300.0 mL of water was added to the solution?

Answer 1

For a: The molality and molarity of the given solution is 2.45m and 2.28 M respectively.

For b: The molarity of the solution when more water is added is 0.912 M

Step-by-step explanation:

• For a:

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

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

Where,

`m_(solute)` = Given mass of solute
`(C_3H_8O_3)` = 42.0 g

`M_(solute)` = Molar mass of solute
`(C_3H_8O_3)` = 92.093 g/mol

`W_(solvent)` = Mass of solvent (water) = 186 g

Putting values in above equation, we get:

`\text{Molality of }C_3H_8O_3=(42* 1000)/(92.093* 186)\\\\\text{Molality of }C_3H_8O_3=2.45m`

• 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)}}` .....(1)

We are given:

Molarity of solution = ?

Molar mass of
`(C_3H_8O_3)` = 92.093 g/mol

Volume of solution = 200 mL

Mass of
`(C_3H_8O_3)` = 42 g

Putting values in above equation, we get:

`\text{Molality of }C_3H_8O_3=(42* 1000)/(92.093* 200)\\\\\text{Molality of }C_3H_8O_3=2.28M`

Hence, the molality and molarity of the given solution is 2.45m and 2.28 M respectively.

• For b:

Now, the 300 mL water is added to the solution. So, the total volume of the solution becomes (200 + 300) = 500 mL

Using equation 1 to calculate the molarity of solution, we get:

Molar mass of
`(C_3H_8O_3)` = 92.093 g/mol

Volume of solution = 500 mL

Mass of
`(C_3H_8O_3)` = 42 g

Putting values in equation 1, we get:

`\text{Molality of }C_3H_8O_3=(42* 1000)/(92.093* 500)\\\\\text{Molality of }C_3H_8O_3=0.912M`

Hence, the molarity of the solution when more water is added is 0.912 M