Search any question & find its solution
Question:
Answered & Verified by Expert
$2.52 \mathrm{~g}$ of oxalic acid dehydrate was disved in $100 \mathrm{ml}$ of water, $10 \mathrm{~mL}$ of this ution was diluted to 500 $\mathrm{mL} .$ The normality of the final ution and the amount of oxalic acid $(\mathrm{mg} / \mathrm{mL}$ ) in the ution are respectively-
Options:
Solution:
1905 Upvotes
Verified Answer
The correct answer is:
$0.04 \mathrm{~N}, 3.60$
Initial Normality
$$
\mathrm{N}=\frac{2.52 \times 1000}{63 \times 100}=0.4
$$
$$
\begin{array}{l}
{ }_{\mathrm{COOH}} \\
\therefore \mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
0.4 \times 10=\mathrm{N}_{2} \times 500 \\
\mathrm{~N}_{2}=\frac{0.4}{50}=0.08 \mathrm{~N}
\end{array}
$$
Then final weight $\quad N=\frac{w \times 1000}{E \times V_{m}}$
$$
0.08=\frac{w \times 1000}{63 \times 500}
$$
$$
\mathrm{N}=\frac{2.52 \times 1000}{63 \times 100}=0.4
$$
$$
\begin{array}{l}
{ }_{\mathrm{COOH}} \\
\therefore \mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
0.4 \times 10=\mathrm{N}_{2} \times 500 \\
\mathrm{~N}_{2}=\frac{0.4}{50}=0.08 \mathrm{~N}
\end{array}
$$
Then final weight $\quad N=\frac{w \times 1000}{E \times V_{m}}$
$$
0.08=\frac{w \times 1000}{63 \times 500}
$$
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.