- A$250\,ml$ solvent
- B$250\,g$ solvent
- ✓$250\,ml$ solution
- D$1000\,ml$ solvent
Molarity $ = \frac{{{\rm{Number\,of \, moles\, of \, solute}}}}{{{\rm{Volume\,of\, solution\, in }}litre}}$
$\therefore 2.0 = \frac{{0.5}}{{{\rm{Volume\, of\, solution\, in}}\,\,litre}}$
Volume of solution in litre $ = \frac{{0.5}}{{2.0}} = 0.250\,litre = 250\,ml.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Which of the following product $(s)$ is/are can be obtained in the above reaction.
$\left(E_{A g^{+} / A g}^{0}=0.80\, V, E_{A n^{+} / A u}^{0}=1.69\, V\right)$
${C_6}{H_5} - N{O_2} + 6[H] \to {C_6}{H_5} - N{H_2} + 2{H_2}O$
The reducing agent used in this reaction is …….
$(CH_3)_2CHCH_2N(CH_2CH_3)_2 \xrightarrow{C{{H}_{3}}I}\xrightarrow[{{H}_{2}}O]{A{{g}_{2}}O}\xrightarrow{heat}$ products