Search any question & find its solution
Question:
Answered & Verified by Expert
The weight of potassium dichromate (molecular weight $=294$ ) required to prepare $0.04 \mathrm{~N}$ of $250 \mathrm{~mL}$ solution is
Options:
Solution:
1746 Upvotes
Verified Answer
The correct answer is:
$0.49 \mathrm{~g}$
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ mostly react in acidic medium as oxidising agent. The reaction occurs as follows :
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+7 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$
$\because$ Reaction involve $6 e^{-}$electrons. Thus,
$\text { Equivalent weight }=\frac{\text { Molar mass }}{\text { Number of electrons }}=\frac{294}{6}$
$=49$
$\because$ We have to prepare $0.04 \mathrm{~N}$ of $250 \mathrm{~mL}$ solution.
$(\because 250 \times 4=1000 \mathrm{~mL})$
Thus, $\frac{0.04}{4}=0.01$ and weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is $=49 \times 0.01=0.49 \mathrm{~g}$
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+7 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$
$\because$ Reaction involve $6 e^{-}$electrons. Thus,
$\text { Equivalent weight }=\frac{\text { Molar mass }}{\text { Number of electrons }}=\frac{294}{6}$
$=49$
$\because$ We have to prepare $0.04 \mathrm{~N}$ of $250 \mathrm{~mL}$ solution.
$(\because 250 \times 4=1000 \mathrm{~mL})$
Thus, $\frac{0.04}{4}=0.01$ and weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is $=49 \times 0.01=0.49 \mathrm{~g}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.