Question

In the presence of excess of anhydrous ${\text{SrC}}{{\text{I}}_2}$,the amount of water taken up is governed by ${{\mathbf{K}}_{\text{p}}} = {10^{12}}\;{\text{at}}{{\text{m}}^{ - 4}}$ for the following reaction at 273K ${\text{SrC}}{{\text{I}}_2} + 2{{\text{H}}_2}{\text{O}}({\text{s}}) + 4{{\text{H}}_2}{\text{O}}({\text{g}}) \leftrightharpoons {\text{SrC}}{{\text{l}}_2} \cdot 6{{\text{H}}_2}{\text{O}}({\text{s}})$. What is equilibrium vapour pressure (in torr) of water in a closed vessel that contains ${\text{SrC}}{{\text{l}}_2} \cdot 2{{\text{H}}_2}{\text{O}}({\text{s}})$ ?
A.0.0001 torr
B.${10^3}$torr
C.0.76 torr
D.0.55 torr

Answer 1

Strontium chloride is a strontium chloride salt. It's a common salt that forms neutral aqueous solutions. This salt, like all Sr compounds, produces a bright red flame; in particular, it is used as a source of redness in fireworks. It has chemical properties that are halfway between those of more toxic barium chloride and less toxic calcium chloride.

Complete answer:
The tendency of a substance to transition into a gaseous or vapour state is measured by vapour pressure, which rises with temperature. The boiling point of a liquid is defined as the temperature at which the vapour pressure at its surface equals the pressure exerted by its surroundings. The evaporation rate of a liquid is determined by the equilibrium vapour pressure. It has to do with particles' proclivity for escaping from liquids (or a solid). Volatile refers to a material that has a high vapour pressure at room temperature.
As the temperature of a liquid rises, so does the kinetic energy of the molecules. The number of molecules transitioning into a vapour grows as the kinetic energy of the molecules increases, raising the vapour pressure. According to the Clausius–Clapeyron relationship, the vapour pressure of any material rises non-linearly with temperature. The temperature at which the vapour pressure matches the ambient air pressure is defined as the atmospheric pressure boiling point of a liquid (also known as the natural boiling point).
${\text{SrC}}{{\text{I}}_2} + 2{{\text{H}}_2}{\text{O}}({\text{s}}) + 4{{\text{H}}_2}{\text{O}}({\text{g}}) \leftrightharpoons {\text{SrC}}{{\text{l}}_2} \cdot 6{{\text{H}}_2}{\text{O}}({\text{s}})$
Number of ${H_2}O$ added in excess is 4
Hence, using Equilibrium vapour pressure formula we get
${K_p} = \dfrac{1}{{P_{{H_2}O(g)}^4}}$
${P_{{H_2}O(g)}} = {\left( {\dfrac{1}{{{K_p}}}} \right)^{1/4}}$
${\left( {{{10}^{ - 17}}} \right)^{1/4}} = {10^{ - 3}}{\text{ atm}}{\text{. }}$
Substituting the given values we get
${K_p} = {10^{ - 3}} \times 760 = 0.76{\text{ torr }}$

Note:
Bubbles forming at a lower level of the liquid necessitates a higher temperature due to the higher fluid pressure, which rises above ambient pressure as depth rises. The higher temperature needed to start bubble forming is more necessary at shallow depths. The bubble wall's surface tension causes an overpressure in the very thin, initial bubbles.