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 Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode  is [at .mass of Cu=63.5 u]

Answer:

Explanation:

Given, Q=2F

  Atomic  mass of Cu= 63.5 u

 Valancy of the metal Z= 2

 we have,   $CuSo_{4}\rightarrow Cu^{2+}+SO_{4}^{2-}$

  $ Cu^{2+}+2e^{-}\rightarrow Cu$

   1 mol     2 mol            1 mol=63.5 g

                 2F

Alternatively,  W= ZQ

                     =  $\frac{E}{F}.2F=2E$

                     =   $\frac{2\times 63.5}{2}=63.5$

The standard  Gibbs energy change at  300K for the reaction,   $2A\rightleftharpoons B+C$  is 2494.2 J.  At a given time , the composition of the reaction mixture is  $[A]=\frac{1}{2}$ , [B]=2   and   $[C]=\frac{1}{2}$ . The reaction proceeds in the  [ R=8.314 JK/mol, e=2.718]

Answer:

Explanation:

 Given,     $\triangle G^{0}=2494.2J$

  $Q=\frac{[B][C]}{[A]^{2}}=\frac{2\times \frac{1}{2}}{(\frac{1}{2})^{2}}=4$

   $\therefore$

  $\triangle G=\triangle G^{0}+RT lnQ$

              =$2494.2+8.314\times 300 ln 4$

             =28747.27  J

 =  Positive value

  Also, we have

          $\triangle G=RT ln\frac{Q}{K}$

 If  $\triangle G$   is positive , Q>K.

 Therefore,   reaction shifts in reverse direction.

 The vapour pressure of acetone at 20° C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20° C, its vapour pressure was 183 torr. The molar mass  (gmol^-1)  of the substance is

Answer:

Explanation:

Given

   P₀   =185 torr at 20° C

   P_s    =183 torr at 20° C

 Mass of non-volatile substance,

                   m=1.2 g

Mass of acetone taken  =100g

              M=  ?

As, we have

           $\frac{P_{0}-P_{s}}{P_{s}}=\frac{n}{N}$

 Putting the values , we get,

  $\frac{185-183}{183}=\frac{\frac{1.2}{M}}{\frac{100}{58}}\Rightarrow\frac{2}{183}=\frac{1.2\times58}{100 \times M}$

$\therefore$                $M=\frac{183\times1.2\times58}{2\times100}$

   M=63.684 =64g/mol

 The following reaction is performed at 298K $2NO(g)+O_{2}(g)\rightleftharpoons 2NO_{2}(g)$. The standard free energy of formation of NO (g)  is 86.6kJ/mol at 298K. What is the standard free energy of formation of NO₂ (g)  at 298K? ($K_{p}=1.6\times 10^{12}$)

Answer:

Explanation:

 For the given reaction,

     $2NO(g)+O_{2}(g)\rightleftharpoons 2NO_{2}(g)$

 Given,   $\triangle G_{f}^{0} (NO)=86.6 kJ/mol$

    $\triangle G_{f}^{0} (NO_{2})=?$

              $K_{p}=1.6\times 10^{12}$

Now , we have

   $\triangle G_{f}^{0}=2\triangle  G_{f(NO_{2})}^{0}-[2\triangle G_{f(NO)}^{0}+ \triangle G_{f(O_{2})}^{0}]$

                  $-RT ln K_{p}=2\triangle  G_{f(NO_{2})}^{0} -(2\times 86600+0)$

   $\triangle  G_{f(NO_{2})}^{0}=\frac{1}{2}$

                                                            $[2\times 86600-R\times298 ln (1.6\times 10^{12}]$

   $\triangle  G_{f(NO_{2})}^{0}=0.5$

                                                   $[2\times 86600-R\times(298) ln (1.6\times 10^{12})]$

The intermolecular interaction that is dependent  on the inverse cube of  distance between the molecules is

Answer:

Explanation:

 Ion-ion interaction is dependent on the square of distance.

  i.e, ion-ion interaction   $\propto  \frac{1}{r^{2}}$

 Similarly,

     ion-dipole  interaction  $\propto  \frac{1}{r^{3}}$

                       London forces  $\propto  \frac{1}{r^{6}}$

  and dipole -dipole interactions    $\propto  \frac{1}{r^{3}}$

 Superficially it seems as both ion dipole interaction and hydrogen bonding vary with inverse cube of distance  between the molecules but when we look at the exact expressions of field (force) created in two situations it comes as

                 $|E| or |F|= \frac{2|P|}{4\pi \epsilon_{0}r^{3}}$

                                  (In case of ion-dipole interaction)

  and               $F=\frac{2q^{2}r-4q^{2}a}{4\pi \epsilon_{0}r^{3}}$

                         ( In case of dipole-dipole interaction)

  From the above it is very clear , the ion-dipole interaction is the better answer as compared to dipole -dipole interaction , i.e hydrogen bonding

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