Answer:
494.49 g of NaCl.
Explanation:
Data obtained from the question include the following:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mass of NaCl =.?
Next, we shall determine the number of mole of NaCl in the solution.
Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as
Molality = mole of solute /Kg of solvent
With the above formula, we can obtain the number of mole NaCl in the solution as follow:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mole of NaCl =..?
Molality = mole of solute /Kg of solvent
3.140 = mole of NaCl /2.692
Cross multiply
Mole of NaCl = 3.140 x 2.692
Mole of NaCl = 8.45288 moles
Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:
Mole of NaCl = 8.45288 moles
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =.?
Mole = mass /Molar mass
8.45288 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 8.45288 × 58.5
Mass of NaCl = 494.49 g.
Therefore, 494.49 g of NaCl are present in the solution.
