JEE 2012 :: General Questions :: Chemistry

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An organic compound undergoes first order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its  initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of 

$\frac{[t_{1/8}]}{[t_{1/10}]}\times10?  (\log_{10} 2=0.3)$

Answer:

Explanation:

The first order process  $kt=ln \frac{[A]_{0}}{[A]}$   where ,$[A]_{0}$=initial concentration

[A]=concentration of reactant remaining at time 't'

$\Rightarrow$    $kt_{1/8}=ln \frac{[A]_{0}}{[A]_{0}/8}=ln 8$..........(i)

and  $kt _{1/10}=ln \frac{[A]_{0}}{[A]_{0}/10}=ln 10$

Therefore,

$\frac{t_{1/8}}{t_{1/10}} =\frac{ln 8}{ln 10}=\log 8=3 \log 2$

                         =$3 \times 0.3=0.9$

$\Rightarrow$  $\frac{t_{1/8}}{t_{1/10}}  \times 10=0.9 \times 10=9$

Which of the following halides react(s) with $AgNO_{3}(aq)$ to give a precipitate that dissolves in $Na_{2}S_{2}O_{3}$ (aq)?

Answer:

Explanation:

Solubilities of silver halides in water decrease from fluoride (AgF) to iodide (AgI). Silver fluoride is readily soluble in water hence when AgNO₃ solution is added to HF solution (HF being weak acid, its solution maintain a very low concentration of F^- ) no precipitate of AgF is formed.

HCl, HBr and HI being all strong acid, forms precipitates of AgCl, AgBr and Agl when AgNO₃ solution is added to their aqueous solution.

$HCl(aq)+AgNO_{3}(aq) \rightarrow AgCl(s)$(Curdy white)+ $HNO_{3}(aq)$

$HBr (aq)+AgNO_{3}(aq) \rightarrow AgBr(s)$(pale yellow)

$HI(aq)+AgNO_{3}(aq) \rightarrow AgI(s)$(yellow)+$HNO_{3}(aq)$

The solubilities decreases from AgCl to AgI, AgCl dissolves in aqueous ammonia, AgBr dissolves only slightly in concentrated ammonia while AgI does not dissolve in ammonia solution $Na_{2}S_{2}O_{3}$ solution dissolve  all toi AgCl,AgBr,AgI by forming compound $[Ag(S_{2}O_{3})_{2}]^{3-}$ as $S_{2}O_{3}^{2-}$ is a strong complexing agent than ammonia

Choose the correct reason(s) for the stability of the lyophobic colloidal particles.

Answer:

Explanation:

Lyophobic sol, which is otherwise unstable, gets stabilized try preferential adsorption of ions on their surface developing a potential difference between the fixed layer and the diffused layer

Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction, as shown in the given scheme.

Answer:

Explanation:

For separation be differential extraction one of the components must form salt with the given base so that the salt will be extracted in an aqueous layer leaving other components in organic layer.

(a) Both phenol and benzoic acid forms salt with NaOH, hence this mixture can't be separated

(b) Benzoic acid forms salt with NaOH while benzyl alcohol does not, hence the mixture can be separated. using NaOH. Also benzoic acid forms salt with NaHCO₃ but benzyl alcohol does not, hence NaHCO₃ can be used for separation.

(c) Neither benzyl alcohol nor phenol forms salt with NaHCO₃, the mixture cannot be separated using NaHCO₃.

(d)$C_{6}H_{5}CH_{2}COOH$forms salt with $NaOH,C_{6}H_{5}CH_{2}OH$  does not, hencemixture can be separated using$NaOH .C_{6}H_{5}CH_{2}COOH$ forms salt  with $NaHCO_{3},C_{6}H_{5}CH_{2}OH$  does not- hence mixture can be separated using $NaHCO_{3}$

For an ideal gas, consider only p-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the
following choice(s) is (are) correct?

Answer:

Explanation:

(a) Entropy is a state function, change tn entropy in a cyclic process is zero'
Therefore.

$\triangle S_{X \rightarrow Y}+\triangle S_{Y \rightarrow Z}+\triangle S_{Z \rightarrow X}$=0

$\Rightarrow$  $-\triangle S_{Z \rightarrow X}=\triangle S_{X \rightarrow Y}+\triangle S_{Y \rightarrow Z}$

=$\triangle S _{X \rightarrow Z}$

Analysis of options (b) and (c)

Work is a non-stable function, it does depend on the Path followed.

$W_{Y\rightarrow Z}=0$ as $\triangle V=0$ Therefore $W_{X\rightarrow Y \rightarrow Z}=W_{X\rightarrow Y}$

Also work is the area under the curve on p-V diagram.

 As shown

 $ W_{X \rightarrow Y}+W_{Y\rightarrow Z}=W_{X\rightarrow Y}= W_{X\rightarrow Y \rightarrow Z}=W_{X\rightarrow Y} $

but not equal to $W_{X \rightarrow Z}$

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