A solution is made by dissolving 14.57 g of sodium bromide (NaBr) in
415 g of water.
What is the molality of the solution?
[?] m NaBr
Molar mass of NaBr: 102.89 g/mol

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.

where are the choice options?