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Answers (1)

According to question,

$\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}$

1                            0                      0

1-$\alpha$                          $\alpha$                    $\alpha$

here $\alpha$ is degree of dissociation.

now,

$\mathrm{K_{a}=\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=\frac{c \alpha \cdot c \alpha}{c(1-\alpha)}=\frac{c \alpha^{2}}{(1-\alpha)}}$

As dissociation constant is very less than 1, so 1-$\alpha$ $\approx$ 1

$\mathrm{\therefore K_{a} = c\alpha ^{2}}$

$\mathrm{\Rightarrow 1.8\times 10^{-5}=c\times \left ( \frac{2}{100} \right )^{2}}$

$\mathrm{c = 0.045\: mole/litre}$

$\text{Amount of acetic acid in 1 litre} \mathrm{= 60 \times 0.045 = 2.7\ g}$

Hence, the correct answer is Option (1)

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