Question

A \[280{\text{ }}mL\] solution is \[20\,\% \] salt. How much water should be added to make the solution \[14\,\% \] salt?

Answer 1

Hint: When one substance dissolves into another substance, forms a solution. A solution is a homogeneous mixture having solute and solvent, in which a solute dissolved into a solvent. The solute is the substance that is being dissolved in a medium that is solvent, while the solvent is the dissolving medium. Solutions can be formed with different types and forms of solutes and solvents. Example, adding sugar in the water where sugar is solute and water is solvent and the mixture of sugar and water is a sugar solution.

Complete step by step solution:In this, we need to determine how much water is added in the \[14\,\% \] of salt solution.
The given values,
Solution \[1\, = \,280{\text{ }}mL\]
Salt percentage \[ = \,20\% \]
Solution \[2\, = \,?\]
Salt percentage \[ = \,14\% \]
Let \[x\] be the grams of salt
Then,
\[{\text{ }}\dfrac{{grams{\text{ }}of{\text{ }}salt}}{{280ml{\text{ }}solution}}\, \times \,100 = 20\% \]
we need \[x\] value so,
\[x\, = \,\dfrac{{280 \times 20}}{{100}}\]
\[x\, = \,56\] grams of salt
Let’s find how much water is added,
This makes it easier to determine how much water to add to the first solution by first calculating how much salt we have.
Consider water as \[y\] ,
So,
\[\dfrac{{56}}{{(280\, + \,y)}}\, \times \,100\, = \,14\% \]
\[\Rightarrow \,56\, \times \,100\, = \,14\,\left( {280\, + \,y} \right)\]
\[\Rightarrow 5600\, = \,3920\, + \,14y\]
\[\Rightarrow 14y\, = \,5600\, - \,3920\]
\[\Rightarrow 14y\, = \,1680\]
\[\Rightarrow y\, = \,\dfrac{{1680}}{{14}}\]
\[\therefore y\, = \,120\,\] litres
Therefore, \[120\,\] litres of water are added to make the solution \[14\% \] salt.

Additional information:Volume percentage:
\[{\text{Percent by volume}}\,{\text{ = }}\,\dfrac{{{\text{Volume}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Volume}}\,{\text{of solution}}}}\,{{ \times }}\,{\text{100}}\,{\text{\% }}\]
The percentage of solute in a solution can be determined by volume when the solute is liquid as well as the solvent is also liquid. The yield of percentage can be shown by dividing the volume of the solute by the volume of solution.
If a solution is made by adding \[20{\text{ }}mL\] of ethanol to \[200{\text{ }}mL\] of water, the percent by volume is;
\[{\text{Percent by volume}}\,{\text{ = }}\,\dfrac{{{\text{Volume}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Volume}}\,{\text{of solution}}}}\,{{ \times }}\,{\text{100}}\,{\text{% }}\]
\[{\text{ = }}\,\dfrac{{20}}{{200}}\,{{ \times }}\,{\text{100}}\,{\text{% }}\]
\[{\text{ = }}\,10\,\% \]

Note: Mass percentage: When the solute in a solution is a solid, a convenient way to express the concentration is a mass percent, which is in the grams of solute per \[100{\text{ }}g\] of solution.
\[{\text{Percent by mass}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Mass}}\,{\text{of solution}}}}\,{{ \times }}\,{\text{100}}\,{\text{% }}\]