Question

Calculate the masses of cane sugar and water required to prepare 250 g of 25% cane sugar solution.

Answer 1

We know that mass percentage is one way to express concentration of a solution using the mass of solute and solution. Here, we have to use the formula of mass percentage, that is, ${\rm{Mass}}\% = \dfrac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solution}}}} \times 100$.

Complete step by step answer:
We know that there are many ways of expressing concentration of a solution, such as, molarity, molality, volume percentage, ppm, mole fraction etc. Here, we have to use a mass percent formula to calculate the mass of solute and solvent.
Given that mass of solution is 250 g and mass percent of the solution is 25%. Now, we have to use the formula of mass percent.
Mass percent of sugar
solution=$\dfrac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{sugar}}}}{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solution} }}} \times 100$
Now, we have to put the values of mass percent of sugar solution (25%) and mass of sugar solution (250 g) in the above equation.
$\Rightarrow 25 = \dfrac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{sugar}}}}{{250}} \times 100$
$\Rightarrow {\rm{Mass}}\,{\rm{of}}\,{\rm{sugar = }}\dfrac{{25 \times 250}}{{100}} = 62.5\,{\rm{g}}$
Therefore, the mass of sugar in the solution is 62.5 g.
Now, we have to calculate the mass of water in the solution. We know that, mass of solute and Solvent gives the total mass of the solution.
The mass of solute (sugar) is 62.5 g and the mass of the solution is 250 g.
Formula is,
Mass of solute (sugar) + Mass of solvent (water)= Mass of solution
Mass of water= Mass of solution- Mass of sugar
$\Rightarrow {\rm{Mass}}\,{\rm{of}}\,{\rm{water}} = 250\,{\rm{g}} - {\rm{62}}{\rm{.5}}\,{\rm{g}} = {\rm{187}}{\rm{.5}}\,{\rm{g}}$

Therefore, the mass of water in the solution is 187.5 g.

Note:
Remember that molarity and molality both are used to measure concentration of a solution. Molarity is the moles of solute present in solution of one litre and molality is the moles of solute present in 1 kg of solvent.