Equivalent weight of K2Cr2O7 in acidic medium is

Question

Equivalent weight of K2Cr2O7 in acidic medium is (A)  M  (B)  (M/2)  (C)  (M/5)  (D)  (M/6)

Tardigrade Admin · Accepted Answer

Sol. K2Cr2O7+4H2SO4 xrightarrow K2SO4+Cr2(SO4)3+4H2O+3[O] or Cr2O72-+14H++6e- xrightarrow2Cr3++7H2O Equivalent weight of K2Cr2O7=(molecular text mass/6)

Q. Equivalent weight of $K_{2} C r_{2} O_{7}$ in acidic medium is

$M$

$\frac{M}{2}$

$\frac{M}{5}$

$\frac{M}{6}$

Sol. $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4}$
$K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 4 H_{2} O + 3 [O]$
or $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} 2 C r^{3 +} + 7 H_{2} O$
Equivalent weight of $K_{2} C r_{2} O_{7} = \frac{m o l ec u l a r ma ss}{6}$

Q. Equivalent weight of K2​Cr2​O7​ in acidic medium is

M

2M​

5M​

6M​

Sol. K2​Cr2​O7​+4H2​SO4​​ K2​SO4​+Cr2​(SO4​)3​+4H2​O+3[O] or Cr2​O72−​+14H++6e−​2Cr3++7H2​O Equivalent weight of K2​Cr2​O7​=6molecularmass​

Q. Equivalent weight of $K_{2} C r_{2} O_{7}$ in acidic medium is

$M$

$\frac{M}{2}$

$\frac{M}{5}$

$\frac{M}{6}$

Sol. $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4}$
$K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 4 H_{2} O + 3 [O]$
or $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} 2 C r^{3 +} + 7 H_{2} O$
Equivalent weight of $K_{2} C r_{2} O_{7} = \frac{m o l ec u l a r ma ss}{6}$