Download App

STD 12 - 8.2 Carboxylic acids and their derivative — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 8.2 Carboxylic acids and their derivative4 Marks
MCQ
$HCOOH$ reacts with conc. $H_2SO_4$ to produce
  • ✓
    $CO$
  • B
    $CO_2$
  • C
    $NO$
  • D
    $NO_2$
✓

Answer

Correct option: A.
$CO$
a
$CO$ $(carbon\,monoxide)$ gas is evolved when a formic acid $( HCOOH )$ reacts with sulphuric acid( $\left.H _2 SO _4\right)$Sulphuric acid is both an oxidizing agent as well as a dehydrating agent.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is
→
If $60\%$ of a first order reaction was completed in $60$ minutes, $50\%$ of the same reaction would be completed in approximately ........... $\min.$

$(\log \,4 = 0.60,\, \log \,5 = 0.69)$

→
During the charging of lead storage Battery the reaction at anode is
→
An alcohol of formula $C_9H_{12}O$ reacts with  $Na_2Cr_2O_7$ to form a  compound having  formula $C_9H_{10}O$. The original alcohol might be 
→
The standard reduction potential for $L{i^ + }/Li\,;\,\,Z{n^{2 + }}/Zn\,;$ ${H^ + }/{H_2}$ and $A{g^ + }/Ag$ is $ - 3.05,\,\, - 0.762,\,\,0.00$ and $ + 0.80\,\,V.$ Which of the following has highest reducing capacity
→
Which dimethylcyclobutane is optically active ?
→
When $100\,\,ml$ of $M/10\,\,NaOH$ solution and $50\,\,ml$ of $M/5\,\,HCl$ solution are mixed, the $pH$ of resulting solution would be
→
Pure methane can be prepared by
→
The electronic configuration of four elements are given in brackets$L\,\left( {1{s^2},\,\,2{s^2}2{p^1}} \right);\,\,M\,\left( {1{s^2},\,\,2{s^2}\,2{p^5}} \right)$

$Q\,\left( {1{s^2},\,\,2{s^2}\,2{p^6},\,\,3{s^1}} \right);\,\,R\,\left( {1{s^2},\,\,2{s^2}\,2{p^2}} \right)$The element that would most readily form a diatomic molecule is

→
In the following reaction, $4P + 3KOH + 3{H_2}O \to 3K{H_2}P{O_2} + P{H_3}$
→