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# The degree of dissociation of $Ca{{(N{{O}_{3}})}_{2}}$ in a dilute aqueous solution containing 7g of the salt per 100g of water at ${{100}^{{}^\circ }}C$ is 70%. If the vapour pressure of water at ${{100}^{{}^\circ }}C$ is 760mm, calculate the vapour pressure of the solution.(A) 378.5mm(B) 492.8mm(C) 746.10mm(D) 985.6mm

Last updated date: 13th Jul 2024
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Hint: The dissociation degree is the fraction of original solute molecules that have dissociated. It is usually indicated by the Greek symbol $\alpha$ . More accurately, the degree of dissociation refers to the amount of solute dissociated into ions or radicals per mole.

Given is the degree of dissociation of $Ca{{(N{{O}_{3}})}_{2}}$, is 70%.
Firstly we will calculate the moles of $Ca{{(N{{O}_{3}})}_{2}}$ and water.
Molar mass of $Ca{{(N{{O}_{3}})}_{2}}$ is 164g/mol and mass of $Ca{{(N{{O}_{3}})}_{2}}$ is 7g. Therefore, moles will be,
$moles=\dfrac{7}{164}=0.0427moles$
Molar mass of water is 18g/mol and mass of water is 100g. Therefore, moles will be,
$moles=\dfrac{100}{18}=5.56moles$
Now the mole fraction (X) of calcium nitrate will be,
$\text{mole fraction=}\dfrac{0.0427}{0.0427+5.56}=0.00762$
The degree of dissociation $\alpha$ =70 % =0.7
The van’t hoff factor will be,
\begin{align} & i=[1+(n-1)\alpha ] \\ & i=[1+(3-1)0.7] \\ & i=2.4 \\ \end{align}
The relative lowering of the vapour pressure, $\dfrac{{{P}^{o}}-P}{{{P}^{o}}}=iX$, where ${{P}^{o}}$ is vapour pressure of the water and P is the vapour pressure of the solution.
Substituting the values of pressure, mole fraction and van’t hoff factor, we get,
\begin{align} & \dfrac{760-P}{760}=2.4\times 0.00762=0.018~ \\ & 760-P=13.9 \\ & P=746.10mm \\ \end{align}

Therefore, the correct answer is the (C) option.

Note: The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.