JEE 2012 :: General Questions :: Chemistry

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For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^{0}C$. Assuming the concentration of solute is much lower than the concentration of solvent, the vapour Pressure (mm of Hg) of the solution is (take  $K_{b}=0.76 K kg mol^{-1})$

Answer:

Explanation:

The elevation in boiling Point is

 $\triangle T_{b}=K_{b},m:m=molality=\frac{n_{2}}{w_{1}} \times 1000$

[$n_{2}$ number of moles of solute,

$w_{1}$ = weight of solvent in gram ]

$\Rightarrow$    $2=0.76 \times \frac{n_{2}}{100}\times 1000$

$\Rightarrow$  $n_{2}=\frac{5}{19}$

Also, from Raoult's law of lowering of
vapour pressure :

$\frac{-\triangle p}{p_{0}}=x_{2}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{n_{2}}{n_{1}}[\because n_{1}>>n_{2}]$

$\Rightarrow$   $-\triangle p=760 \times \frac{5}{19} \times \frac{18}{100}$

=36 mm of Hg

 $\Rightarrow$     p=760-36=724 mm of Hg

The shape of $XeO_{2}F_{2}$ molecule is 

Answer:

Explanation:

In  $XeO_{2}F_{2}$  the bonding arrangement around the central atom Xe is 

 $\Rightarrow$  Hybridization of Xe=$sp^{3}d$

$sp^{3}d$  hybridization corresponds to trigonal bipyramidal geometry. Also, in trigonal bipyramidal geometry, lone pairs remain present on equatorial positions in order to give less electronic repulsion.

.

Note According to Bent's ru1e, the more electronegative atoms must be present on the axial position. Hence, F is kept on axial positions.

Using the data provided, calculate the multiple bond energy (kJ mol^-1)of a $C\equiv C$ bond $C_{2}H_{2}$. That energy is (take the bond energy of a C-H bond as 350k Jmol-1)

$2C(s)+H_{2}(g) \rightarrow   C_{2}H_{2}(g);$     $\triangle H=225kJ mol^{-1}$

$2C(s) \rightarrow  2 C (g)$      $\triangle H=1410 kJ mol^{-1}$

$H_{2}(g) \rightarrow 2H(g)$   $\triangle H=330 kJ Mol^{-1}$

Answer:

Explanation:

For the calaculation of $C\equiv C$  bond energy we must first calculate dissociation energy of $C_{2}H_{2}$ as

 $C_{2}H_{2}$(g)$ \rightarrow$ $2C(g)+2H(g)$.....(i)

 Using the given bond energoes and enthalpies

 $C_{2}H_{2}(g) \rightarrow 2C(g)+2H(g);$

                             $\triangle H=-225kJ$ ........(ii)

$2C (s) \rightarrow 2C(g);$ $ \triangle H=1410kJ$..........(iii)

 $H_{2}(g) \rightarrow 2H(g);$ $\triangle H=330kJ$......(iv)

 adding equation (ii),(iii) and (iv) gives equation(i)

 $\Rightarrow$      $C_{2}H_{2}(g) \rightarrow 2C(g)+2H(g)$;

                                       $\triangle H=1515 kJ$

 $\Rightarrow$     1515kJ=$2 \times$ (C-H)BE+($C\equiv C$)BE

 =$2 \times 350$+($C\equiv C$)BE

$\Rightarrow$     $(C\equiv C)$BE=1515-700=815 kJ/mol

In the cyanide extraction process of silver from argentite ore' the oxidizing and reducing agents used are

Answer:

Explanation:

The reactions involved in the extraction of silver by the cyanide process are

   $Ag_{2}S+CN^{-}+O_{2}  \rightarrow   [Ag(CN)_{2}]^{-}+SO_{2}$ ..................(i)

$[Ag(CN)_{2}]^{-}+Zn \rightarrow [Zn(CN)_{4}]^{2-}$+Ag ..........(ii)

In reaction (i), sulphide is oxidized to SO₂ by oxygen. In the reaction (ii),

Silver ion ($Ag^{+}$) is reduced to Ag by Zn.

Therefore,  $O_{2}$ . is oxidisinS agent and Zn is reducing agent.

The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorous containing compound. The reaction type; the oxidation  states of phosphrous in phosphine and the other product are respectively.

Answer:

Explanation:

The reaction of white phosphorus with aqueous alkali is

 $P_{4}+3NaOH+3H_{2}O \rightarrow  PH_{3}+NaH_{2}PO_{2}$

In the above reaction phosphorus is simultaneously oxidized

$[P_{4}(O) \rightarrow NaH_{2} P^{+1}O_{2}]$  as well as reduced  [$P_{4}(0) \rightarrow PH_{3}]$ 

Therefore, this is an example of disproportionation reaction. Oxidation number of phosphorus in $PH_{3}$ is -3 and in $NaH_{2}PO_{2}$ is +1.

However,+1 oxidation no. is not given in any option, one might think that $NaH_{2}PO_{2}$, has gone to further decomposition on heating.

 $2NaH_{2}PO_{2}$ $\underrightarrow{\triangle} Na_{2}H^{+5}PO_{4}+PH_{3}$

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