JEE 2012 :: General Questions :: Chemistry

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With respect to graphite and diamond, which of the statement(s) given below is (are) correct

Answer:

Explanation:

Diamond has a three-dimensional network- structure, a hard substance where graphite is soft due to layered structure. In graphite, only three valence electrons are involved in bonding and one electron remains free giving electrical conductivity. ln diamond, all the four valence electrons are covalently bonded hence, insulator.

Diamond ls a better thermal conductor than graphite. Electrical conductivity is due to the availability of free electrons, thermal conduction is due to transfer of thermal vibrational energy from one atom to another atom. Compact and precisely aligned crystals like diamond thus facilitate better movement of heat

In graphite C-C bond acquire some double-bond character, hence, higher bond order than in diamond.

Bleaching Powder and bleach solutions are Produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid, is

Answer:

Explanation:

Bleaching powder is $Ca(OCl)Cl.$ Therefore, the oxoacid whose salt is present in bleaching powder is HOCl. Anhydride of  HOCl is $-Cl_{2}O$ as $2HOCl \rightarrow Cl_{2}O+H_{2}O$

[Note The oxidation number of element in anhydride and oxoacid remains the same]

Bleaching Powder and bleach solutions are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N $Na_{2}S_{2}O_{3}$, was used to reach the endpoint. The molarity of the household bleach solution is

Answer:

Explanation:

The involved redox reactions are :

 $2H^{+}+OCl^{-}+2I^{-} \rightarrow Cl^{-}+I_{2}+H_{2}O$ ....(i)

$I_{2}+2S_{2}O_{3}^{2-} \rightarrow 2I^{-}+S_{4}O_{6}^{2-}$ ......(ii)

 Also   the n factor of$S-{2}O_{3}^{2-}$ is one as

 $2S_{2}O_{3}^{2-} \rightarrow S_{4}O_{3}^{2-}+2e^{-}$

[obe 'e' is produced per unit of $S_{2}O_{3}^{2-}$]

$\Rightarrow$  Molarity of $Na_{2}S_{2}O_{3}=0.25 N \times 1$0.25 M

$\Rightarrow$  m mol of $Na_{2}S_{2}O_{3}$ used up=$0.25 \times 48=12$

Now from stoichiometry of reaction (ii)

12 m mol of $S_{2}O_{3}^{2-}$ would have reduced 6mmol of $I_{2}$

From stoichiometry of reaction (i)

m mol of $OCl^{-}$ reduced =m mol of $I_{2}$ produced =6

 $\Rightarrow$ Molarity of household  bleach solution= $\frac{6}{25}=0.24 M$

The electrochemical cell shown below is a concentration cell $M|M^{2+}$ (saturated solution of a sparingly  soluble salt, $MX_{2})$|| $M^{2+}$ (0.001 mol dm^-3)|M. The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of  the cell at 298 is 0.059 V. The value of $\triangle G$ (kJ mol^-1) for the given cell is (take 1F=96500C mol^-1)

Answer:

Explanation:

For the given concentration of the cell, the cell reactions are $M \rightarrow M^{2+}$ at left hand electrode

$M^{2+} \rightarrow M$ at right hand electrode

$\Rightarrow$ $M^{2+}$(RHS electrode)$

$\rightarrow M^{2+}$ (LHS elelctrode)

$E^{0}=0$

Applying Nernst equation

$E_{cell}=0.059=0-\frac{0.059}{2}$

$\log\frac{[M^{2+}]at LHS electrode}{0.001}$
$\Rightarrow$ $\log\frac{[M^{2+}]at LHS electrode}{0.001}=-2$
$\Rightarrow$ $[M^{2+}]$ at LHS electrode

=$10^{-2} \times 0.001=10^{-5} M$

$\triangle G=-nEF=-\frac{2 \times 0.059 \times 96500}{1000} kJ$

 =$-11.4 KJ$

The electrochemical cell shown below is a concentration cell $M|M^{2+}$ (saturated solution of a sparingly  soluble salt, $MX_{2})$|| $M^{2+}$ (0.001 mol dm^-3)|M. The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of  the cell at 298 is 0.059 V. The solubility product $ (k_{sp};mol^{3}dm^{-9})$  of MX, at 298 based on the information available the given concentration cell is (take 2.303 x R x 298/F= 0.059 V)

Answer:

Explanation:

For the given centration of the cell, the cell reactions are  $M \rightarrow M^{2+}$ at left hand electrode

$M^{2+} \rightarrow  M$  at right hand electrode

$\Rightarrow$ $M^{2+}$(RHS elecrode)

                     $\rightarrow  M^{2+}$ (LHS elelctrode)

$E^{0}=0$

Applying Nernst equation

 $E_{cell}=0.059=0-\frac{0.059}{2}$

$\log\frac{[M^{2+}]at LHS electrode}{0.001}$

$\Rightarrow$   $\log\frac{[M^{2+}]at LHS electrode}{0.001}=-2$

$\Rightarrow$  $[M^{2+}]$ at LHS electrode 

   =$10^{-2} \times 0.001=10^{-5} M$

The solubility equilibria for  $MX_{2}$ is 

$MX_{2}\rightleftarrows M^{2+}(aq)+2X^{-}(aq)$

 Solubility product , $K_{sp}=[M^{2+}][X^{-}]^{2}$

=  $10^{-5} \times (2 \times 10^{-5})^{2}=4 \times 10^{-15}$

  [$ \because  $  In satuarated solution of $MX_{2}$,$[X^{-}]$=2$[M^{2}]]$

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