Question

40g of MgO dissolves in \[{H_2}O\] to form \[200ml\] of a solution, if the density of the solution is \[1.5g/ml\]. Calculate ppm of solute.

Answer 1

Hint: To solve the above numerical we should know the basic concepts and terms of chemistry. We should be able to determine the solute and solvent in the solution. Here we are using the measure of concentration as ppm.

Formula Used:
Ppm of solute \[ = \dfrac{{w\left( {mg} \right)}}{{V\left( L \right)}}\]
Where \[w\left( {mg} \right)\] is the amount of solute in milligram and \[V\left( L \right)\] is the volume of solution in liters.

Complete step by step answer:
First, we will understand the basic terms used in chemistry which are solute and solvent. The solute is the substance that is dissolved in the solution. Whereas the solvent is the part of the solution in which the solute is dissolved. Both the solute and the solvent from the solution. Now we will concentrate on the given quantities in the question. So here we have given that \[40g\] \[MgO\] dissolves in \[{H_2}O\] to form \[200ml\] of solution. Now we can conclude that some amount of \[MgO\] is dissolved in\[{H_2}O\]. Therefore, \[MgO\] is the solute and \[{H_2}O\] is solvent.
So now in the question, we have given that \[w = 40g\] and \[V = 200ml\]. Now we will substitute the given quantities in the above formula by converting them into the required units. After converting the given quantities into the required units we get \[w = 40 \times {10^3}mg,V = 200 \times {10^{ - 3}}L\]. Now we will substitute in the formula,
Ppm of solute \[ = \dfrac{{w\left( {mg} \right)}}{{V\left( L \right)}} = \dfrac{{40 \times {{10}^3}}}{{200 \times {{10}^{ - 3}}}}\]
\[ \Rightarrow \] ppm of solute \[ = 2 \times {10^5}mg/L\]

Therefore, the calculated value of ppm of solute which is dissolved in the solvent forming a \[200ml\] solution is \[2 \times {10^5}mg/L\].

Note: We can also calculate the molarity of the solution from the given quantities by calculating the number of moles of solute using the molecular weight and given mass of the solute. The volume of the solution is already given.