Question

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, 181 grams of MgCl2 are recovered. What is the molarity of MgCl2 in the original sample?

Answer 1

Answer:

1.90 M

Explanation:

Step 1: Given data

• Mass of MgCl₂ (solute): 181 g

• Volume of the solution: 1.00 L

Step 2: Calculate the moles corresponding to 181 g of MgCl₂

The molar mass of MgCl₂ is 95.21 g/mol.

181 g × 1 mol/95.21 g = 1.90 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 1.90 mol/1.00 L = 1.90 M