When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier than ordinary seawater) is evaporated, 151 grams of MgCl2 are recovered. What is the molarity of MgCl2 in the original sample?

1.59mol/L

Explanation:

Data obtained from the question include:

Mass of MgCl2 = 151g

Volume of water(solvent) = 1L

Now, let us calculate the number of mole of MgCl2. This is illustrated below:

Molarity Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g/mol

Mass of MgCl2 = 151g

Number of mole of MgCl2 =?

Number of mole = Mass /Molar Mass

Number of mole of MgCl2 = 151/95

Number of mole of MgCl2 = 1.59mole

Now we can calculate the molarity of MgCl2 as follow:

Mole = 1.59mole

Volume = 1L

Molarity =?

Molarity = mole /Volume

Molarity = 1.59/1

Molarity = 1.59mol/L

a is the correct answer