Question

If you had a salt-water aquarium that needs salt water, could you figure out how much sea salt to add to address a salt deficit? Assume that salt is the only solute in the seawater.

You may need:

average ocean salinity = 35g/L

Average density of ocean = 1.025g/ml

1L = 0.26 gal

a) What is the Molarity, M, of the ocean?

b) What is the molality, m, of the ocean?

c) What is the mole fraction, X, of salt in the ocean?

d) How much salt in kg would be required for a 55-gallon tank?

Answer 1

Salt is sodium chloride, NaCl, having a molar mass of (1*22.9897 + 1*35.453) g/mol = 58.4427 g/mol.

a) The salinity of ocean water is 35 g/L, i.e, 1 L of ocean water contains 35 g salt.

Mole(s) of salt corresponding to 35 g = (35 g)/(58.4427 g/mol) = 0.5989 mol.

Molarity of ocean water = (mole(s) of salt)/(volume of ocean water) = (0.5989 mol)/(1 L) = 0.5989 mol/L ≈ 0.60 M (ans).

b) The density of ocean water is 1.025 g/mL; therefore, mass of 1 L of ocean water = (1 L)*(1000 mL/1 L)*(1.025 g/mL) = 1025 g = (1025 g)*(1 kg/1000 g) = 1.025 kg.

Molality of ocean water = (mole(s) of salt)/(kilograms of ocean water) = (0.5989 mole)/(1.025 kg) = 0.58429 mol/kg ≈ 0.58m (ans).

c) Molar mass of water = (2*1.008 + 1*15.9994) g/mol = 18.0154 g/mol.

Moles of ocean water corresponding to 1025 g = (1025 g)/(18.0154 g/mol) = 56.8958 mole.

Mole fraction of salt in ocean water, X = (mole(s)of salt)/(total number of moles of salt and water) = (0.5989 mole)/[(0.5989 + 56.8958) mole] = 0.0104 (ans).

d) 1 L = 0.26 gallon; therefore, 55 gallon = (55 gallon)*(1 L/0.26 gallon) = 211.5385 L.

1 L of ocean water = 35 g salt; therefore,

211.5385 L ocean water = (35 g/1 L)*(211.5385 L) = 7403.8475 g salt = (7403.8475 g)*(1 kg/1000 g) = 7.40384 kg ≈ 7.404 kg (ans).