Equivalent weight of C6H5CHO is equal to molar mass in the following reaction. Thus, species (A) is

Question

Equivalent weight of C6H5CHO is equal to molar mass in the following reaction. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c07ec6301309c5894c081a7f15e3d2ad-.jpeg" /> Thus, species (A) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Gain of one O-atom

≡1(O)≡16 g≡8×2 g(O) (M/2)≡8g(O) Thus, E (oxidation part) =(M/2) Since, E (net) = M Hence, E (reduction part) = (M/2)

∴ E (reduction) =(M/2) Hence, A is

CH2OH

Q. Equivalent weight of $C_{6} H_{5} C H O$ is equal to molar mass in the following reaction.

Thus, species (A) is

Gain of one O-atom
$\equiv 1 (O) \equiv 16 g \equiv 8 \times 2 g (O) \frac{M}{2} \equiv 8 g (O)$
Thus, E (oxidation part) = $\frac{M}{2}$
Since, E (net) = M
Hence, E (reduction part) = $\frac{M}{2}$

$∴$ E (reduction) = $\frac{M}{2}$
Hence, A is $C H_{2} O H$

Q. Equivalent weight of C6​H5​CHO is equal to molar mass in the following reaction. Thus, species (A) is

Gain of one O-atom ≡1(O)≡16g≡8×2g(O)2M​≡8g(O) Thus, E (oxidation part) =2M​ Since, E (net) = M Hence, E (reduction part) = 2M​ ∴ E (reduction) =2M​ Hence, A is CH2​OH

Q. Equivalent weight of $C_{6} H_{5} C H O$ is equal to molar mass in the following reaction.

Thus, species (A) is

Gain of one O-atom
$\equiv 1 (O) \equiv 16 g \equiv 8 \times 2 g (O) \frac{M}{2} \equiv 8 g (O)$
Thus, E (oxidation part) = $\frac{M}{2}$
Since, E (net) = M
Hence, E (reduction part) = $\frac{M}{2}$

$∴$ E (reduction) = $\frac{M}{2}$
Hence, A is $C H_{2} O H$