ChemistryQ1–2 of 2 questions
Page 1 of 1 · English
1
ChemistryEasyMCQGSEB · 2015
What would the boiling point elevation of $0.1 \ m$ $KNO_3$ be equal to?
A
$0.1 \ m$ Urea
B
$0.1 \ m$ sodium chloride
C
$0.1 \ m$ potassium sulfate
D
$0.1 \ m$ aluminum nitrate
Solution
(B) The boiling point elevation is given by the formula $\Delta T_b = i \times K_b \times m$. Since $K_b$ and $m$ are constant for the given solutions,the boiling point elevation depends on the van't Hoff factor $(i)$.
For $KNO_3$,it dissociates as $K^+ + NO_3^-$,so $i = 2$.
For $0.1 \ m$ Urea,$i = 1$ (non-electrolyte).
For $0.1 \ m$ sodium chloride $(NaCl)$,it dissociates as $Na^+ + Cl^-$,so $i = 2$.
For $0.1 \ m$ potassium sulfate $(K_2SO_4)$,it dissociates as $2K^+ + SO_4^{2-}$,so $i = 3$.
For $0.1 \ m$ aluminum nitrate $(Al(NO_3)_3)$,it dissociates as $Al^{3+} + 3NO_3^-$,so $i = 4$.
Therefore,the boiling point elevation of $0.1 \ m$ $KNO_3$ is equal to that of $0.1 \ m$ sodium chloride.
For $KNO_3$,it dissociates as $K^+ + NO_3^-$,so $i = 2$.
For $0.1 \ m$ Urea,$i = 1$ (non-electrolyte).
For $0.1 \ m$ sodium chloride $(NaCl)$,it dissociates as $Na^+ + Cl^-$,so $i = 2$.
For $0.1 \ m$ potassium sulfate $(K_2SO_4)$,it dissociates as $2K^+ + SO_4^{2-}$,so $i = 3$.
For $0.1 \ m$ aluminum nitrate $(Al(NO_3)_3)$,it dissociates as $Al^{3+} + 3NO_3^-$,so $i = 4$.
Therefore,the boiling point elevation of $0.1 \ m$ $KNO_3$ is equal to that of $0.1 \ m$ sodium chloride.
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2
ChemistryEasyMCQGSEB · 2015
If $1$ mole of a solute dissociates to give $n$ number of ions,then the degree of dissociation $\alpha$ in terms of the van't Hoff factor $i$ is given by:
A
$\frac{i-1}{n+1}$
B
$\frac{n-1}{i-1}$
C
$\frac{i-1}{n-1}$
D
$\frac{n+1}{i-1}$
Solution
(C) For a solute dissociating as $A \rightarrow nB$,the initial moles are $1$ and $0$ respectively.
After dissociation,the moles are $(1-\alpha)$ and $n\alpha$.
The total number of moles at equilibrium is $(1-\alpha + n\alpha) = 1 + \alpha(n-1)$.
The van't Hoff factor $i$ is defined as the ratio of observed moles to initial moles: $i = \frac{1 + \alpha(n-1)}{1}$.
Rearranging for $\alpha$: $i - 1 = \alpha(n-1)$.
Therefore,$\alpha = \frac{i-1}{n-1}$.
After dissociation,the moles are $(1-\alpha)$ and $n\alpha$.
The total number of moles at equilibrium is $(1-\alpha + n\alpha) = 1 + \alpha(n-1)$.
The van't Hoff factor $i$ is defined as the ratio of observed moles to initial moles: $i = \frac{1 + \alpha(n-1)}{1}$.
Rearranging for $\alpha$: $i - 1 = \alpha(n-1)$.
Therefore,$\alpha = \frac{i-1}{n-1}$.
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