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

Step-by-step explanation:

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

Formula used :

`Molarity=(n* 1000)/(V_s)`

where,

n= moles of solute

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}=(0.049g)/(95g/mol)=5.2* 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=(5.2* 10^(-4)moles* 1000)/(100ml)=5.2* 10^(-3)mole/L`

Therefore, the molarity of solution will be
`5.2* 10^(-3)mole/L`

`MgCl_2\rightarrow Mg^(2+)+2Cl^-`

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

Thus
`5.2* 10^(-3)` moles of
`MgCl_2` gives =
`(2)/(1)* 5.2* 10^(-3)=0.0104`

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