Answer:
0.342 m
Explanation:
From the question given above, the following data were obtained:
Mass of NaBr = 14.57 g
Mass of water = 415 g
Molar mass of NaBr = 102.89 g/mol
Molality of NaBr =?
Next, we shall determine the number of mole in 14.57 g of NaBr. This can be obtained as follow:
Mass of NaBr = 14.57 g
Molar mass of NaBr = 102.89 g/mol
Mole of NaBr =?
Mole = mass / molar mass
Mole of NaBr = 14.57 / 102.89
Mole of NaBr = 0.142 mole
Next, we shall convert 415 g of water to kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
415 g = 415 g × 1 Kg / 1000 g
415 g = 0.415 Kg
Thus, 415 g is equivalent to 0.415 Kg.
Finally, we shall determine Molality of the solution as follow:
Mole of NaBr = 0.142 mole
Mass of water = 0.415 Kg
Molality of NaBr =?
Molality = mole / mass of water in Kg
Molality of NaBr = 0.142 / 0.415
Molality of NaBr = 0.342 m
Therefore, the molality of NaBr solution is 0.342 m.
