Question

A solution was prepared by dissolving 40.0 g of kcl in 225 g of water. part a calculate the mass percent of kcl in the solution. express your answer with the appropriate units. 15.1 {\rm \\%} calculate the mole fraction of the ionic species kcl in the solution.

Answer 1

 part A :

The mass % of KCl = 17.78 %

calculation

 mass%= mass  of the solute/mass of the solvent x100

mass of the solute  40 g

mass of the solvent=225

% mass= 40/225 x100=17.78%

Part B

The mole fraction of ionic species =0.04296

the mole  fraction=moles of KCl/ moles of the solution

moles of KCl=  40 g/74.5 g/mol=0.537 moles

moles of water  = 225g/18 g/mol=12.5 moles

mole fraction is therefore= 0.537/12.5=0.04296

Answer 2

Explanation:

(a)   According to the given data, mass of the solution will be calculated as follows.

          Mass of solution = Mass of KCl + Mass of water

                                      = (40 + 225) g

                                      = 265 g

Therefore, (m/m)% of KCl will be calculated as follows.

           (m/m)% of KCl = $\frac{\text{mass of KCl}}{\text{mass of solution}} \times 100$    

                                   = $\frac{40}{265} \times 100$

                                   = 15.09%

Hence, mass percent of KCl in the solution is 15.09%.

(b)    Now, we will calculate the mole fraction of ionic species present in KCl  in the solution are follows.

Molar mass of KCl is 74.55 g/mol.

                Moles of KCl = $\frac{mass}{\text{molar mass}}$

                                      = $\frac{40 g}{74.55 g/mol}$

                                      = 0.536 mol

or,                                   = 0.54 mol

As, molar mass of water is 18 g/mol. Hence, moles of water will be calculated as follows.

           Moles of $H_{2}O$ = $\frac{mass}{\text{molar mass}}$

                                      = $\frac{225 g}{18 g/mol}$

                                      = 12.5 mol

So, total number of moles = moles of KCl + moles of $H_{2}O$

                                           = (0.54 + 12.5) mol

                                           = 13.04 mol

Calculate the mole fraction of KCl in the solution as follows.

              Mole fraction = $\frac{\text{moles of KCl}}{\text{total moles}}$

                                     = $\frac{0.54 mol}{13.04 mol}$

                                     = 0.041

Therefore, mole fraction of the ionic species KCl in the solution is 0.041.