A) Volume strength of remaining \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] solution will be \[\frac{\text{78}\text{.4}}{\text{12}}\text{V}\].
B) Molarity of \[{{\text{I}}_{\text{2}}}\]in solution X is 0.025 M
C) Molarity of hypo solution taken is 0.2 M
D) Moles of tetrathionate ions formed will be 0.01
Correct Answer: D
Solution :
\[{{M}_{{{H}_{2}}{{O}_{2}}}}=\frac{V.S.}{11.2}\Rightarrow {{M}_{{{H}_{2}}{{O}_{2}}}}=1M\] \[{{n}_{{{H}_{2}}{{O}_{2}}}}=1\times 0.1=0.1\] \[{{n}_{KI}}=0.5\times 0.05=0.025\] \[\underset{0.1}{\mathop{{{H}_{2}}{{O}_{2}}+}}\,\]\[\underset{0.0.025}{\mathop{2KI}}\,\xrightarrow[{}]{{}}\underset{0}{\mathop{{{I}_{2}}}}\,\]\[+\,\underset{0}{\mathop{2KOH}}\,\]? \[\left( 0.1-\frac{0.025}{2} \right)0\]\[\frac{0.025}{2}\] \[0.025\] \[{{n}_{I2}}=\frac{0.025}{2}\]in 500 ml \[\therefore \]\[{{M}_{{{I}_{2}}}}=2\times \frac{0.025}{2}=0.025M\] Remaining moles of \[{{H}_{2}}{{O}_{2}}=0.1-\frac{0.025}{2}=\frac{0.175}{2}\]in 150 ml \[\therefore \]\[M=\frac{0.175}{2}\times \frac{1000}{150}=\frac{{{(V.S.)}_{{{H}_{2}}{{O}_{2}}}}}{11.2}\] \[\therefore \]\[V.S{{.}_{{{H}_{2}}{{O}_{2}}}}=\frac{0.175}{2}\times \frac{1000}{150}\times 11.2=\frac{78.4}{12}\] \[2{{S}_{2}}O_{3}^{2-}+{{I}_{2}}\xrightarrow[{}]{{}}{{S}_{4}}O_{6}^{2-}+2{{I}^{-}}\] \[M\times 0.05(0.025\times 0.2)\] \[\therefore \]\[M\times 0.05=2\times 0.025\times 0.2\] \[{{M}_{{{S}_{2}}O_{3}^{2-}}}=0.2\] \[\therefore \]Molarity of hypo solution = 0. 2 M Moles of tetrathionate ions \[=0.025\times 0.2=0.005\]molesYou need to login to perform this action.
You will be redirected in
3 sec