JEE 2014 :: Main 2014 :: Chemistry 2014

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For the complete combustion of ethanol.   $C_{2}H_{5}OH(l)+3O_{2}(g)\rightarrow3CO_{2}(g)+3H_{2}O(l)$, the amount of heat  produced  as measured in bomb calorimeter, is 1364.47 kJ mol^-1  at 25° C. Assuring ideality the enthalpy of combustion, $\triangle_{C}H$, for the reaction  will be (R=8.314 JK^-1mol^-1)

Answer:

Explanation:

$C_{2}H_{5}OH(l)+3O_{2}(g)\rightarrow3CO_{2}(g)+3H_{2}O(l)$ 

$\triangle_{}U=-1364.47 kJ/mol$

$\triangle_{}H=\triangle_{}U+\triangle n_{g }RT$

 $\triangle n_{g }$  = -1

  $\triangle H=-1364.47+\frac{-1\times8.314\times298}{1000}$

 [Here, value of R in unit of J must be converted into kJ]

 =-1364.47-2.4776

 =-1366.94 kJ/mol

 During solving such problem, students are advised to keep much importance in unit conversion. As here, value of R 

(8.314 Jk^-1 mol^-1) in Jk^-1 mol^-1 must be converted into kJ by dividing the unit by 1000.

 Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution of 0.5M solution of same electrolyte is 1.4 Sm^-1 and resistance of same solution of the same electrolyte is 280Ω. The molar conductivity of 0.5 M solution  of the electrolyte in Sm² mol^-1 is 

Answer:

Explanation:

 In order to solve the problem, calculate the value of cell constant of the first solution and then use this value of cell constant to calculate the value of k of second solution. Afterwards, finally, calculate molar conductivity using value of k and m.

 For first solution,

  k=1.4 Sm^-1 , R=50Ω, M= 0.2

 Specific conductance (k)=   $\frac{1}{R}\times\frac{1}{A}$

$1.4 Sm^{-1}=\frac{1}{50}\times\frac{1}{A}$

$\frac{1}{A}=50\times 1.4 m^{-1}$

  for second solution. 

          R=280,   $\frac{1}{A}=50\times 1.4 m^{-1}$

  $k=\frac{1}{280}\times50\times 1.4 =\frac{1}{4}$

 Now, molar conductivity

   $\lambda_{m}=\frac{K}{1000\times m}= \frac{1}{2000}=5\times 10^{-4}Sm^{2}mol^{-1}$

For the esimation  of nitrogen, 1.4 g of an organic compound was digested  by Kjeldahl's  method and the evolved ammonia  was absorbed in 60 mL  of   $\frac{M}{10}$ sulphuric acid. The unreacted  acid required 20 mL of   $\frac{M}{10}$ sodium hydroxide  for complete  neutralisation . The percentage  of nitrogen in the compound

Answer:

Explanation:

 This problem is based on the estimation of percentage of N in organic compound using Kjeldahl's method. Use the concept of stoichiometry  and follow the steps given below to solve the problem

 (i)  Write the balanced chemical reaction for the conversion of N present in organic compound to ammonia, ammonia to ammonia sulphate and ammonium sulphate to sodium sulphate.

 (ii) Calculate  millimoles (m moles ) of N present in organic compound followed by mass of N present in organic compound using the concept of stoichiometry

(iii) At last calculate  % of N present  in organic compound using formula

 %  of N=  Mass of N X 100/ mass of organic compound.

Mass of organic compound =1.4 g

 Let it contain x mmole of N atom.

  Organic compound →  NH₃ (x mmole)

$ 2NH_{3}+H_{2}SO_{4}\rightarrow (NH_{4})_{2}SO_{4}$        ............(i)

                6m mole initially taken

$H_{2}SO_{4}+2NaOH\rightarrow Na_{2}SO_{4}+2H_{2}O$   .............(ii)

 2mmoles NaOH reacted.

 Hence mmoles of   $H_{2}SO_{4}$ reacted inEq.(ii)=1

$\Rightarrow$ mmoles of   $H_{2}SO_{4}$ reacted in Eq.(i)=6-1=5 mmoles

$\Rightarrow$   mmoles of   $NH_{3}$ in Eq.(i)= 2 x 5= 10.mmoles

$\Rightarrow$ mmoles of N atom  in the organic compound=10mmoles

  $\Rightarrow$  Mass of N=   $10\times 10^{-3}\times14=0.14 g$

 % of N= Mass of N present in organic compound/ Mass of organic compound  x 100

$\Rightarrow$       % of N=   $\frac{0.14}{1.4}\times100=10$%

CsCl crystallises in body centred cubic lattice. If a its edge length, then which of the following expression is correct?

Answer:

Explanation:

In CsCl,Cl^- lie at  corners of simple cube and Cs⁺ at the body centre . Hence  , along the  body diagonal , Cs⁺ and Cl^- touch each other so,

 $r_{Cs^{+}}+r_{Cl^{-}}=2r$

 Calculation of   $\tau$

In $\triangle EDF$

Body centred cubic unit cell

   $FD=b=\sqrt{a^{2}+a^{2}}=\sqrt{2}a$

  In  $\triangle AFD$, 

          $c^{2}=a^{2}+b^{2}$

               $=a^{2}+(\sqrt{2}a)^{2}$

   $=a^{2}+2a^{2}$

         $c^{2}=3a^{2}$

         $c=\sqrt{3}a$

As  $\triangle AFD$ is an equilateral triangle,

  $\therefore$     $\sqrt{3}a=4r$

  $\Rightarrow   r =\frac{\sqrt{3}a}{4}$

  Hence,   $r_{Cs^{+}}+r_{Cl^{-}}=2r$

                =  $2\times\frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{2}a$

If Z is a compressibility  factor, van der Waal's equation at low pressure can be written as 

Answer:

Explanation:

To solve this problem the stepwise approach required i.e,

 (i) write the van der Waal's equation, then apply the condition that  at low pressure, volume become high,

    i.e, V-b=V

 Now, calculate  the value of  compressibility factor (Z). [Z=pV/RT]

 According to Van der Waal's equation,

                    $\left(p+\frac{a}{V^{2}}\right)(V-b)=RT$

at low pressure,

     $\left(p+\frac{a}{V^{2}}\right)V=RT$

$pV+\frac{a}{V}=RT$

$pV=RT-\frac{a}{V}$

Divide both sid e by RT

        $\frac{pV}{RT}=1-\frac{a}{RTV}$

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014