Question

What are the respective concentrations (M) of Mg2+and Cl-afforded by dissolving 0.325 mol MgCl2 in water and diluting to 250 mL?A) 0.0013 and 0.0013B) 0.0013 and 0.0026C) 1.30 and.1.30D) 1.30 and 2.60E) 2.60 and 2.60

Answer 1

Answer: D) 1.30 and 2.60

Explanation:

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

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

molarity  of $MgCl_2$ = $\frac{0.325\times 1000}{250ml}=1.30 M$

$MgCl_2\rightarrow Mg^{2+}+Cl^{-}$

As 1 mole of $MgCl_2$ dissociates to give one mole of $Mg^{2+}$ and two moles of  $Cl^-$ ions .

Molarity of $Mg^{2+}$ = 1.30 M

Molarity of $Cl^{-}$ = $1.3 M\times 2=2.60M$