Question

How much water must be evaporated from 60 kilograms of a 15% (m/m) salt solution in order to obtain a 40% solution?

Answer 1

It is given that the initial strength of the solution is 15% (mass/mass). So, with the help of this we can calculate the total mass of salt and water present in solution and then we can further calculate the mass of water required to get evaporated

Complete answer:
Let us see, what we are given in the question:-
Initial strength of the given solution is 15% (m/m)
Which means the salt present in 60 kg solution is = $60kg\times \dfrac{15}{100}$ = 9kg
Therefore, the water present in the solution = (60-9) kg = 51kg
When we evaporate some water off the solution, the salt does not change its mass, unless it decomposes or reacts.
Let's assume that we have evaporated ‘x’ kg of water. Then,
According to the given condition:-
$\dfrac{\text{salt present in the solution}}{\text{water left in the solution after evaporation}}\times 100=desired\text{ }solution$
Water left in the solution after evaporation = (51-x) kg
On substituting all the values, we get:-
=$\dfrac{9kg}{(51-x)kg}\times 100=40$
=$\dfrac{900}{(51-x)}=40$
=$\dfrac{900}{40}=(51-x)$
=$22.5=(51-x)$
= x = 28.5kg
Which means 28.5 kg of water needs to be evaporated to obtain a 40% solution.

Additional Information:
There are many concentration terms and few of them are given below:-
-Mass by volume percentage: Amount of solute (in gram) present in 100mL of solution.
($\dfrac{\text{mass of solute}}{\text{volume of solution}}\times 100$ )
-Mass by mass percentage: Amount of solute (in gram) present in 100g of solution. ($\dfrac{\text{mass of solute}}{\text{mass of solution}}\times 100$ )
-Volume by volume percentage: volume of solute present in 100mL solution.($\dfrac{\text{volume of solute}}{\text{volume of solution}}\times 100$ )

Note:
Kindly remember to accordingly convert units and use them in the formula as required to get the correct result.