MCQ
The empirical formula of a compound is $C{H_2}O.$ $0.0835$ moles of the compound contains $1.0\ g$ of hydrogen. Molecular formula of the compound is
- ✓${C_6}{H_{12}}{O_6}$
- B${C_5}{H_{10}}{O_5}$
- C${C_4}{H_8}{O_8}$
- D${C_3}{H_6}{O_3}$
$\therefore $ $1\,gm$ mole of compound contain = $\frac{1}{{0.0835}} = 11.97$ $=12\,gm$ of hydrogen.
$12\, gm$ of ${H_2}$ is present in ${C_6}{H_{12}}{O_6}$
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(Given ${\Delta _{fus}}H = 6\, kJ\, mol^{-1}$ at $0\,^oC$,
$C_p(H_2O, l) =75.3\, J\, mol^{-1} \, K^{-1}$ ,
$C_p(H_2O, s) = 36.8\, J\, mol^{-1} \, K^{ -1}$ )
$C_{(graphite)} + CO_{2(g)} \rightarrow 2CO _{(g)}$
are $170\, kJ$ and $170\, J K^{-1},$ respectively. This reaction will be spontaneous at ............ $\mathrm{K}$