Question

In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water should you add?

Answer 1

Answer: The mass of water that should be added in 203.07 grams

Step-by-step explanation:

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

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

Where,

m = molality of barium iodide solution = 0.175 m

`m_(solute)` = Given mass of solute (barium iodide) = 13.9 g

`M_(solute)` = Molar mass of solute (barium iodide) = 391.14 g/mol

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

Putting values in above equation, we get:

`0.175=(13.9* 1000)/(391.14* W_(solvent))\\\\W_(solvent)=(13.9* 1000)/(391.14* 0.175)=203.07g`

Hence, the mass of water that should be added in 203.07 grams