Answer: The mass of water that should be added in 203.07 grams
Explanation:
To calculate the molality of solution, we use the equation:
[tex]\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
Where,
m = molality of barium iodide solution = 0.175 m
[tex]m_{solute}[/tex] = Given mass of solute (barium iodide) = 13.9 g
[tex]M_{solute}[/tex] = Molar mass of solute (barium iodide) = 391.14 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = ? g
Putting values in above equation, we get:
[tex]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[/tex]
Hence, the mass of water that should be added in 203.07 grams
