Question

A chemist prepares a solution of magnesium chloride MgCl2 by measuring out 49.mg of MgCl2 into a 100.mL volumetric flask and filling to the mark with distilled water.Calculate the molarity of Cl−anions in the chemist's solution.Be sure your answer is rounded to the correct number of significant digits.

Answer 1

Answer:  Molarity of $Cl^-$ anions in the chemist's solution is 0.0104 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

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

where,

n= moles of solute

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.049g}{95g/mol}=5.2\times 10^{-4}moles$  

$V_s$ = volume of solution in ml = 100 ml

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

$Molarity=\frac{5.2\times 10^{-4}moles\times 1000}{100ml}=5.2\times 10^{-3}mole/L$

Therefore, the molarity of solution will be $5.2\times 10^{-3}mole/L$

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

As 1 mole of $MgCl_2$ gives 2 moles of $Cl^-$

Thus $5.2\times 10^{-3}$  moles of $MgCl_2$ gives =$\frac{2}{1}\times 5.2\times 10^{-3}=0.0104$

Thus the molarity of $Cl^-$ anions in the chemist's solution is 0.0104 M