VITEEE 2015 :: Chemistry :: General questions

• Chemistry - General questions

 Given,  $\triangle G^{0}=-nFE^{0}_{cell}$ and  $\triangle G^{0}=-RT In k$  .The value of n=2 will be given by the slope of which line in the figure

Answer:

Explanation:

As we know that $-nFE^{0}_{cell}=-RT$ In k  or   $E^{0}_{cell}=\frac{RT}{nF}In k$

 plot of In k or $E^{0}_{cell}$ will have  slope $=\frac{1}{2}\frac{RT}{F}$

Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction.

 $C_{6}H_{6}(g)+3H_{2}(g)\rightarrow C_{6}H_{12}(g)$

A mixture of C₆H₆ and excess H₂ has a pressure of 60mm of Hg in an unknown volume. After the gas had been passed over a nickel catalyst and all the benzene converted to cyclohexane ,the pressure of the gas was 30mm of Hg in the same volume at the same temperature. The fraction of C₆H₆ (by volume) present in the original  volume is 

Answer:

Explanation:

Let initially ,pressure of C₆H₆ (g) ids  p₁   mm and for H₂(g) is p₂ mm

 p₁+p₂= 60 mm ....(i)

 After the reaction pressure of C₆H₆(g)=0 (as all C₆H₆  has reacted)

 H₂(g)= p₂-3p₁

 So, total pressure =  $p_{2}-3p_{1}+p_{1}=30$mm

 $p_{2}-2p_{1}=30$mm ..........(ii)

 On solving equation (i) and (ii)

 $p_{1}$=10mm, $p_{2}$=50mm

 Fraction of C₆H₆ by volume= moles fraction of pressure =  $\frac{10}{60}=\frac{1}{6}$

 An alloy  of copper , silver and gold is found to have copper atom constiuting the ccp lattice. If silver atom occupy the edge centres and gold atom is present at body centred, the alloy has a formula 

Answer:

Explanation:

 In the unit  cell number of Cu atoms (fcc/ccp)

= $8\times\frac{1}{6}+6\times\frac{1}{2}=4$

 As Ag atoms occupying edge centred = $12\times\frac{1}{4}=3$

 and Au  atoms are presents at the body centred=1

 $\therefore$  formula , $Cu_{4}Ag_{3}Au$

A certain  metal when irradiated by light (r=3.2 x 10¹⁶ Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (r=2.0 x 10¹⁶ Hz)

The v₀ of metal is 

Answer:

Explanation:

(kE)₁ = hv₁- hv₀

 (kE)₂= hv₂-hv₀

 As, (KE)₁ =2 x (KE)₂

$\therefore$    $hv_{1}-hv_{0}=2(hv_{2}-hv_{0})$

or    $hv_{0}=2hv_{2}-hv_{1}$

 or   $v_{0}=2v_{2}-v_{1}$

 $=2\times (2\times 10^{16})-(3.2\times 10^{16})$

 $=0.8\times 10^{16}Hz=8\times 10^{15}Hz$

Amongest 

$NO_{3}^{-},AsO_{3}^{3-},CO_{3}^{2-},ClO_{3}^{-},SO_{3}^{2-}$ and $BO_{3}^{3-}$

 the non-plannar species are 

Answer:

Explanation:

$AsO_{3}^{3-} ,ClO_{3}^{-}$ and $SO_{3}^{2-}$   have sp² hybridisation and hence are non- planar species, while 

$NO_{3}^{-} , CO_{3}^{2-} $  and $ BO_{3}^{3-}$ have sp² hybridisation and hence are planar species

• Chemistry - General questions