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

Explanation:

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

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

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:

$0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g$

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