Question

How many millilitres of a $ 2.5{\text{M}} MgC{l_2} $ solution contain $ 17.5{\text{g}} MgC{l_2}? $

Answer 1

Hint: To determine how many moles present in the given mass, you would have in that sample, use the molar mass of magnesium chloride. Use the molarity of the solution to find out how many litres that many moles would contain. And then we will finally convert the volume to millilitres from litres.

Formula used:
We would require the formula for molarity to solve this question
$ {\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}} $
Where, $ {n_s} $ is the number of moles of solute
${v_s} $ is the volume of solvent in litres.

Complete step by step solution:
According to the question,
Molarity of $ MgC{l_2} $ solution $ = 2.5 M $
Molecular weight of $ MgC{l_2} $ $ = 95.21 {\text{g/mol}} $
This would mean that every mole of $ MgC{l_2} $ will have a mass of $ 95.21 {\text{g/mol}} $
We have to find what volume of $ 2.5{\text{M}} MgC{l_2} $ would contain $ 17.5{\text{g}} MgC{l_2} $
First let us find out the number of moles in $ 17.5{\text{g}} MgC{l_2} $
This can be calculated by simple unitary method
$ 95.21{\text{ g}} $ has $ 1 $ mole of $ MgC{l_2} $
So, $ 17.5{\text{g}} $ will have $ \dfrac{1}{{95.21}} \times 17.5 = 0.18 $ moles of $ MgC{l_2} $
Now, we will calculate the volume of the solution with the help of formula above
$ {\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}} $
Rearranging the formula will give us the equation of volume
$ {v_s} = \dfrac{{{n_s}}}{{{\text{molarity}}}} $
Now, let us put the known values in the formula
$ {v_s} = \dfrac{{0.18 {\text{moles}}}}{{2.5 {\text{M}}}} = 0.07352 {\text{L}} $
Now, we will convert litre into millilitres to get or required answer
Hence, the required volume is $ 0.07352 {\text{L}} = 73.52{\text{ mL}} $.

Additional information:
Molarity, which is referred to as molar, is represented by M. The molarity of a solution where one gram of solute is dissolved in a litre of solution is one molar. As we know, the solvent and solute blend is used in a solution to form a solution, so the total volume of the solution is taken.

Note:
In solutions, the majority of reactions occur and it is therefore essential to understand how the amount of substance is expressed when it is present in the solution. The quantity of substances in the solution is expressed in many ways.