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Question 14 Marks
Read the following text carefully and answer the questions that follow:
Raoult's law for volatile liquids states that the partial vapour pressure of each component in the solution is directly proportional to its mole fraction, whereas for a non $-$ volatile solute, it states that the vapour pressure of a solution of a non $-$ volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction. Two liquids $A$ and $B$ are mixed with each other to form a solution, the vapour phase consists of both components of the solution. Once the components in the solution have reached equilibrium, the total vapour pressure of the solution can be determined by combining Raoult's law with Dalton's law of partial pressures. If a non $-$ volatile solute $B$ is dissolved into a solvent $A$ to form a solution, the vapour pressure of the solution will be lower than that of the pure solvent. The solutions which obey Raoult's law over the entire range of concentration are ideal solutions, whereas the solutions for which vapour pressure is either higher or lowerthan that predicted by Raoult's law are called non $-$ ideal solutions. Non $-$ ideal solutions are identified by determining the strength of the intermolecular forces between the different molecules in that particular solution.They can either show positive or negative deviation from Raoult's law depending on whether the $A - B$ interactions in solution are stronger or weaker than $A - A$ and $B - B$ interactions.
$i. 20 \ mL$ of a liquid $A$ was mixed with $20 \ mL$ of liquid $B$. The volume of resulting solution was found to be less than $40 \ mL$. What do you conclude from the above data?
$ii$. Which of the following show positive deviation from Raoult's law? Carbon disulphide and Acetone; Phenol and Aniline; Ethanol and Acetone.
$iii$. The vapour pressure of a solution of glucose in water is $750\ mm\ Hg$ at $100^\circ C$. Calculate the mole fraction of solute.
$($Vapour pressure of water at $373 K = 760 \ mm\ Hg)$
OR
$iii$. The boiling point of solution increases when $1$ mol of $\ce{NaCl}$ is added to $1$ litre of water while addition of $1$ mol of methanol to one litre of water decreases its boiling point. Explain the above observations.
Answer
$i$. Solution shows a negative deviation from Raoult's law 
$A-A $ and $B-B$ interactions are weaker than $A-B$ interactions.
$ii.$ Carbon disulphide and acetone, Ethanol and acetone.
$iii.$  According to Raoult's law:
$P_1=p_1^0 x_1 $ or $ x_1=\frac{p_1}{p_1^0}$
$x_1=\frac{750}{760}=0.987$
$x_2=1-x_1$
$=1-0.987=0.013$
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Question 24 Marks
Read the following text carefully and answer the questions that follow:
$KMnO _4$ and $K _2 Cr _2 O _7$ are most important chemicals which are used as oxidising agents and disinfectants. $K _2 MnO _4$ is prepared by fusing $MnO _2$ with KOH in presence of $O _2 \cdot K_2 MnO _4$ is electrolysed to get purple coloured $KMnO _4 \cdot Na _2 CrO _4$ is prepared by heating chromite ore with $Na _2 CO _3$ in presence of $O _2 \cdot Na _2 CrO _4$ is converted into $Na _2 Cr _2 O _7$ by reacting with concentrated $H _2 SO _4 \cdot Na _2 Cr _2 O _7$ is reacted with KCl to get $K _2 Cr _2 O _7$, orange coloured solid, soluble in water, changes to yellow coloured $CrO _4^{2-}$ in basic medium, $KMnO _4$ acts as oxidising agent in acidic, neutral as well basic medium. In acidic medium, it converts $Fe ^{2+}$ to $Fe ^{3+}, Sn ^{2+}$ to $Sn ^{4+}, COO ^{-}$to $CO ^2$. In basic medium it converts $I ^{-}$to $IO ^{3-} . K _2 Cr _2 O _7$ acts as oxidising agent only in acidic medium, converts $H _2 S$ to $S , SO _2$ to $SO ^{2-}, I ^{-}$to $I _2$. Lanthanoids and actinoids belong to f-block elements with general electronic configuration $(n-2) f^1$ to $14(n-1) d^{0-2} n s^2$. All actinoids are radioactive. Both show contraction in atomic and ionic radii but actinoid contraction is more than lanthanoid contraction. Lanthanoid show +3 oxidation state, few elements show +2 and +4 oxidation states also. Actinoids show $+3,+4,+5,+6,+7$ oxidation states.
i. Which lanthanoid shows +4 oxidation state and why?
ii. Give two similarities between lanthanoids and actinoids.
iii. Complete the equation and balance:
$
Cr_2 O_7^{2-}+Fe^{2+}+H^{+} \rightarrow ?
$
OR
iii. Convert sodium chromate to sodium dichromate. Give chemical equation.
$2 Na _2 CrO _4+ H _2 SO _4$ (conc.) $\rightarrow$ ?
Answer
i. 'Ce' shows +4 oxidation state because it has stable noble gas electronic configuration.
ii. i. Both show contraction, lanthanoid and actinoid contraction.
ii. Both form-coloured ions and undergo f-f transition.
iii. $Cr _2 O _7^{2-}+6 Fe ^{2+}+14 H ^{+} \rightarrow 2 Cr ^{3+}+7 H _2 O +6 Fe ^{3+}$
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