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

Answer 1

Answer: Molarity of $MgCl_2$ in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{\text{no of moles}}{\text{Volume in L}}$

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.96}{1.00L}$

$Molarity=1.96mol/L$

Thus molarity of $MgCl_2$ in the original sample was 1.96M