Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

In the standardization of $Nas_2S_2O_3$ using $K_2Cr_2O_7$ by iodometry , the equivalent weight of $K_2Cr_2O_7$ is

$(a)\;\text{(molecular weight)/2} \\ (b)\;\text{(molecular weight )/6} \\(c)\;\text{(molecular weight)/3} \\(d)\;\text{Same as molecular weight} $

1 Answer

Comment
A)
In iodometry , $K_2Cr_2O_7$ acts as an oxidising agent to liberate iodine $(I_2)$ from an iodide such as $KI$ or $NaI$
The iodine $(I_2) $ thus liberated. is titrated with hypo $(Na_2S_2O_3)$ solution.
$Na_2S_2O_3+I_2 \longrightarrow Na_2S_4O_6 +2 NaI$
We find that one mole of $K_2Cr_2O_7$ accepts 6 moles of electron as follows.
$Cr_2O_7^{2-}+14H^{+}+6e^{-} \longrightarrow 2Cr{3+}+7H_2O$
Thus equivalent weight $=\large\frac{Molecular \;weight}{6}$
Hence b is the correct answer.
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...