91Ó°ÊÓ

Molarity:

Molarity is oneof the metrics used to express the concentration of a solution.

Molarity = the number of moles of solute / the volume of a solution in L.

Molality:

Molality is one of the numerous factors used to express a solution's concentration.

Molality= the number of moles of solute / the mass of a solvent in kg.

Mass Percent:

Mass percent is one of the metrics used to express the concentration of a solution.

Mass percent of a solute = (the mass of solute / the mass of the solution)$\mathbf{×}\mathbf{100}\mathbf{\%}$

Consider that the volume of the solution is 1.000 L, or 1.15 kg, because the density of the solution is 1.15 g / mLThe mass of HF used to make a 49 percent HF solution is as follows:

The mass percent of HF = (the mass of HF / the mass of the solution)$×100\%$

Rearrange the expression to get:

The mass of HF = the mass percent x the mass of the solution

Replace the values of the mass percent and the mass of the solution.

The mass of HF is, therefore, calculated as:

$$\begin{gathered} =49\%×1.15kg \\ =0.49x1.15kg \\ =0.5635kgHF. \end{gathered}$$

Now, the mass of the solution = the mass of the solute + the mass of the solvent

= the mass of HF + the mass of the water.

The mass of the solute, HF is calculated as

So, 1.15 kg of solution $=0.5635kgofHF+themassofthewater.$

= 1.15 kg- 0.5635 kg

= 0.5865 kg

Next, the number of moles of Hydrogen fluoride is:

The no. of moles of HF = the mass of HF / the molar mass of HF

$$\begin{gathered} =563.5g/20.01g/mol \\ =28.2mol \end{gathered}$$

The no. of moles of water = the mass of Hâ‚‚O / the molar mass of Hâ‚‚O

$$\begin{gathered} =0.5865g/18.02g/mol \\ =0.0325mol. \end{gathered}$$

The molality of the solution is calculated as:

The molality of HF= the number of moles of HF / the mass of Hâ‚‚O in kg

$$\begin{gathered} =28.2mol/0.5865kg \\ =48m. \end{gathered}$$

Thus, the molality is 48m

The molarity of the solution is calculated as:

The molarity of HF = the number of moles HF / the volume of the solution in L.

$$\begin{gathered} =28.2mol/1.000L \\ =28.2M \end{gathered}$$

So, the molarity is 28.2M