229153
An oxygen containing organic compound was found to contain $52 \%$ carbon and $13 \%$ of hydrogen. Its vapour density is 23 . The compound reacts with sodium metal to liberate hydrogen. A functional isomer of this compound is
1 Ethanal
2 Methoxy methane
3 Methoxy ethane
4 Ethanol
Explanation:
Molecular mass $=2 \times$ vapour density Molecular mass $=2 \times 23=46$ {|c|c|c|c|c|} | Element | Percentage | ${l} { Atomic } | |---|---|---| | { mass }$ | ${l} { Re lative } | | { number of } | | { atoms }$ | ${l} { Simplest } | | { ratio }$ | |$$ | $52 \%$ | 12 | $52 / 12=4.33$ | $1.9-2$ | |$$ | $13 \%$ | 1 | $13 / 1=13$ | $5.9-6$ | |$$ | $35 \%$ | 16 | $35 / 16=2.18$ | 1 | | Empirical formula $=\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ Empirical formula mass $=12 \times 2+6 \times 1+16=46$ $\mathrm{n}=\frac{\text { Empirical formula mass }}{\text { Molecular formula mass }}=\frac{46}{46}=1$ Molecular formula $=\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ Alcohols liberate hydrogen on reaction with sodium metal. The functional isomer of ethanol is methoxy methane. $\underset{\text { Ethanol }}{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}} \quad \underset{\text { Methoxy methane}}{\mathrm{CH}_{3} \mathrm{OCH}_{3}}$
Karnataka-CET-2012
Chemical Equilibrium
229152
On increasing temperature, the equilibrium constant of exothermic and endothermic reactions, respectively :
1 Increases and decreases
2 Decreases and increases
3 Increases and increases
4 Decreases and decreases
Explanation:
$A+B \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} C+D$ $\therefore \quad$ Equilibrium constant $(\mathrm{K})=\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}$ For exothermic reaction $\mathrm{K}_{\mathrm{f}}$ decrease with increase of temperature so that $\mathrm{K}$ decreases and endothermic reaction $\left(\mathrm{K}_{\mathrm{f}}\right)$ increases with increase of temperature. So $\mathrm{K}$ increases.
229153
An oxygen containing organic compound was found to contain $52 \%$ carbon and $13 \%$ of hydrogen. Its vapour density is 23 . The compound reacts with sodium metal to liberate hydrogen. A functional isomer of this compound is
1 Ethanal
2 Methoxy methane
3 Methoxy ethane
4 Ethanol
Explanation:
Molecular mass $=2 \times$ vapour density Molecular mass $=2 \times 23=46$ {|c|c|c|c|c|} | Element | Percentage | ${l} { Atomic } | |---|---|---| | { mass }$ | ${l} { Re lative } | | { number of } | | { atoms }$ | ${l} { Simplest } | | { ratio }$ | |$$ | $52 \%$ | 12 | $52 / 12=4.33$ | $1.9-2$ | |$$ | $13 \%$ | 1 | $13 / 1=13$ | $5.9-6$ | |$$ | $35 \%$ | 16 | $35 / 16=2.18$ | 1 | | Empirical formula $=\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ Empirical formula mass $=12 \times 2+6 \times 1+16=46$ $\mathrm{n}=\frac{\text { Empirical formula mass }}{\text { Molecular formula mass }}=\frac{46}{46}=1$ Molecular formula $=\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ Alcohols liberate hydrogen on reaction with sodium metal. The functional isomer of ethanol is methoxy methane. $\underset{\text { Ethanol }}{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}} \quad \underset{\text { Methoxy methane}}{\mathrm{CH}_{3} \mathrm{OCH}_{3}}$
Karnataka-CET-2012
Chemical Equilibrium
229152
On increasing temperature, the equilibrium constant of exothermic and endothermic reactions, respectively :
1 Increases and decreases
2 Decreases and increases
3 Increases and increases
4 Decreases and decreases
Explanation:
$A+B \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} C+D$ $\therefore \quad$ Equilibrium constant $(\mathrm{K})=\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}$ For exothermic reaction $\mathrm{K}_{\mathrm{f}}$ decrease with increase of temperature so that $\mathrm{K}$ decreases and endothermic reaction $\left(\mathrm{K}_{\mathrm{f}}\right)$ increases with increase of temperature. So $\mathrm{K}$ increases.