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Given ,$E^{0}_{Cl_{2}/Cl^{-}}$=1.36 V

  $E^{0}_{Cr^{3+}/Cr}=-0.74 V$

$E^{0}_{Cr_{2}O^{2-}_{7}/Cr^{3+}}=1.33 V$

$E^{0}_{MnO^{-}_{4}/Mn^{2+}}=1.51 V$

Among the following , the strongest reducing agent is

Answer:

Explanation:

The substance which has the lowest reduction potentials are stronger reducing agents. Therefore, Cr ( $E^{0}_{Cr^{3+}/Cr}=-0.74 V$ ) is the strongest reducing agent among the other given options.

1 g of a carbonate (M₂CO₃) on treatment with excess HCl produces 0.01186 mole of CO₂ . The molar mass M₂CO₃ in g mol^-1 is

Answer:

Explanation:

$M_{2}CO_{3}+2HCl\rightarrow 2MCl+H_{2}O+CO_{2}$

1g                                                                  0.01186 mole

Number of moles of M₂CO₃ reacted = Number of moles of CO₂ evolved

$\frac{1}{M}=$  0.01186 [M = Molar mass of M₂CO₃]

$M=\frac{1}{0.01186}=84.3 g mol^{-1}$

A water sample has ppm level concentration of following  anions $F^{-} =10 $:  $SO_{4}^{2-} =100 $ :  $NO_{3}^{-}=50$ the anion /anions that, make/makes the water sample unsuitable for drinking is/are

Answer:

Explanation:

$NO_{3}^{-}$  The maximum limit of nitrate (NO₃^-) in drinking water is 50 ppm and its source is fertilizers. If the maximum limit is increased in water, it will cause methemoglobinemia (blue baby syndrome)

$SO_{4}^{2-} $   The maximum limit of sulphate ($SO_{4}^{2-} $) according to WHO is 500ppm and its source is acid rain, industries. Excess $SO_{4}^{2-} $ has laxative effect.

 $F^{-} $   The maximum limit of fluoride ( $F^{-} $)is about 1.5 ppm its highest concentration converts enamel to more hardest fluorapatite.  Concentration (>2 ppm) causes brown mottling of teeth and high concentration (>10ppm) is harmful to bones and teeth.

$\therefore$    $SO_{4}^{2-} $ (100 ppm) and  $NO_{3}^{-}$ (50 ppm) in water is suitable for drinking but the concentration of $F^{-} $ (10 ppm)makes water unsuitable for drinking.

On the treatment of 100 mL of 0.1 M solution of CoCl₃. 6H₂O with an excess of AgNO₃: 1.2× 10²² ions are precipitated. The complex is

Answer:

Explanation:

Molarity (M)

            = Number of moles of solute/Volume of solution (in L)

$\therefore$    number of moles of complex=  $\frac{Molarity \times volume ( mL)}{1000}$

               = $\frac{0.1 \times 100}{1000}=0.01 mole$

   Number of  moles of ions precipitate= $\frac{1.2 \times 10^{22}}{6.02 \times10^{23}}=0.02 mole$

$\therefore$   Number of Cl^- present in ionisation sphere

       = Number of moles of ions precipitated/ Number of moles of complex

                              = $\frac{0.02}{0.01}=2$

$\therefore$ Cl^- are present outside the square brackets. i.e. in ionisation sphere .Thus the formula of complex is 

[Co(H₂O)₅Cl] Cl₂ .H₂O

The freezing point of benzene decreases by 0.45° C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a  dimer in benzene, percentage association of acetic acid in benzene will be (K_f for benzene=5.12 K kg mol^-1 )

Answer:

Explanation:

Let the degree of association of acetic acid  (CH₃COOH) in benzene is $\alpha$  then

                   $2CH_{3}COOH\rightleftarrows (CH_{3}COOH)_{2}$

initial moles                         1                                   0

Moles at equilibrium           1- $\alpha$                      $\frac{\alpha}{2}$

$\therefore$    Total moles = $1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$

    or                         i $=1-\frac{\alpha}{2}$

 Now, depression in freezing point (Δ T_f )  is given as

       Δ T_f = i K_f m                 ...........(i)

where, K_f = molal depression constant or cryoscopic constant

m=molality

   Molality = number of moles of solute / weight of solvent (in kg )

 $=\frac{0.2}{60}\times\frac{1000}{20}$

  Putting the values in Eq. (i)

$\therefore$    $0.45 =[1-\frac{\alpha}{2}](5.12)[\frac{0.2}{60}\times\frac{1000}{20}]$

$1-\frac{\alpha}{2}= \frac{0.45\times 60\times 20}{5.12\times0.2\times1000}$

$\Rightarrow$  $1-\frac{\alpha}{2}$ =0.527 $\Rightarrow$    $\frac{\alpha}{2}=1-0.527$

     $\therefore$      $\alpha$  =0.946

               The percentage of association  =94.6 %

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