Question

Constant-boiling aqueous HCl can be used as a primary standard for acid-base titrations. When,20 wt% HCl (FM 36.461) is distilled, the composition of the distillate varies in a regular manner with the barometric pressure:

(a) Make a graph of the data in the table to find the weight percent of HCl collected at 746 Torr.

(b) What mass of distillate (weighed in air, using weights whose density is 8.0 g/mL) should be dissolved in 1.000 0 L to give 0.100 00 M HCl? The density of distillate over the whole range in the table is close to 1.096

g/mL. You will need this density to change the mass measured in vacuum to mass measured in air. See Section 2-3 for buoyancy corrections.

Short answer

(a) The weight percent of$\mathrm{HCI}$at pressure 746 torr is$20.254\%$.

(b) 17.99 g of $\mathrm{HCI}$distillate has to be dissolved into 1.000 L to give$0.10000\mathrm{MHCI}$.

Step-by-step solution

Define molarity

Molarity is one of the parameters used to express the concentration of a solution. It is expressed as,

$\mathbf{Molarity}\mathbf{}\mathbf{=}\frac{\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}\mathbf{}\mathbf{in}\mathbf{}\mathbf{L}}$

a) Plot graph for the data in the table

A graph is plotted using the above-given data is,

To calculate the weight percent of $\mathrm{HCI}$at 746 Torr, we can substitute the value of $\mathrm{y}$with 746:

$$\begin{gathered} \mathrm{y}=-416.67\mathrm{x}+9185746 \\ =-416.67\mathrm{x}+9185416.67\mathrm{x} \\ =8439\mathrm{x} \\ =\mathrm{wt}\left(\mathrm{HCI}\right) \\ =20.254\% \end{gathered}$$

From the graph above weight percent of $\mathrm{HCI}$at pressure 746 torr is$20.254\%$.

b) Determine the mass

$0.10000\mathrm{mol}\mathrm{of}\mathrm{HCI}$dissolved in $1.000\mathrm{L}$of water gives $0.10000\mathrm{M}\mathrm{HCI}$ $0.10000\mathrm{mol}\mathrm{of}\mathrm{HCI}$is equivalent to$0.10000\mathrm{molx}36.461\mathrm{g}-\mathrm{mol}=3.6461\mathrm{g}.$Therefore, the solution required

$\frac{3.6461\mathrm{g}\mathrm{HCI}}{0.20254\mathrm{gHCI}/\mathrm{g}\mathrm{solution}}=18.002\mathrm{gsolution}$

Therefore, the mass required

$$\begin{gathered} \mathrm{m}=\frac{18.002\mathrm{x}\left(1-{\displaystyle \frac{0.0012}{1.096}}\right)}{\left(1-\frac{0.0012}{8.0}\right)} \\ =17.99\mathrm{g} \end{gathered}$$