Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The vapor pressure of water at${{23}^{\circ }}C$is $\text{19}\cdot \text{8mm}$. $\text{0}\text{.1}$glucose is dissolved in $178\cdot 2g$ water. What is the vapor pressure (in mm) of the resultant solution?A. 19B. 19.602C. 19.402D. 19.202

Last updated date: 19th Sep 2024
Total views: 413.4k
Views today: 4.13k
Verified
413.4k+ views
Hint: Vapor pressure of a solution can be found using Raoult’s Law. According to this law, the partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.

Formula used:$p={{p}_{o}}{{x}_{A}}$
$p\to$vapor pressure of solution
${{p}_{0}}\to$vapor pressure of pure component
${{\text{x}}_{A}}\to \text{ }$mole fraction
${{\text{x}}_{A}}=\dfrac{{{n}_{A}}}{{{n}_{B}}+{{n}_{B}}}$
Where ${{n}_{A}}\to$no. of moles of pure component (water)
Where ${{n}_{B}}\to$no. of moles of glucose

Vapor Pressure: It is a measure of the tendency of a material to change into gaseous or vapor state.
Mole fraction: It is defined as the unit of amount of constituents divided by the total amount of all constituents in a mixture.
Given,
${{n}_{B}}$(number of moles of glucose)$=0\cdot 1$
${{n}_{B}}$(number of moles of water)$=\dfrac{\text{Given weight of substance}}{\text{Molar mass}}$
Molar mass for water$\left( {{\text{H}}_{2}}\text{O} \right)=2\times 1$(for${{\text{H}}_{2}})+16$(for O)
$=2+16=18$
So,
${{n}_{A}}=\dfrac{178\cdot 2}{18}=9\cdot 9$
${{x}_{A}}=\dfrac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}=\dfrac{9.9}{9\cdot 9+0\cdot 1}=\dfrac{9.9}{10\cdot 0}=0\cdot 99$

\begin{align} & \text{So, vapour pressure of solution (p)=}{{\text{p}}_{0}}{{x}_{A}} \\ & \\ \end{align}
$=19\cdot 8\times 0\cdot 99$
$=19\cdot 602mm$ .

So, the correct answer is “Option B”.