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The molar solubility of $Zn(OH)_2 $ ....

The molar solubility of $Zn(OH)_2 $ in 0.1 M NaOH solution is $x \times 10^{-18} M$. The value of x is _ _ _ _ . (Nearest integer)

(Given: The solubility product of $Zn(OH)_2 $ is $2 \times 10^{-20} $)

Solution

$Zn{(OH)_2}\rightleftharpoons \mathop {Z{n^{2 + }}}\limits_s  + 2\mathop {O{H^ - }}\limits_{2s + 0.1} $

${K_{sp}} = [Z{n^{2 + }}]{[O{H^ - }]^2} = s.{(2s + 0.1)^2}$

$\therefore 2 \times {10^{ - 20}} = s.{(2s + 0.1)^2} \approx {0.1^2} \times s$

$ \Rightarrow s = 2 \times {10^{ - 18}}$

$\therefore x = 2 $

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$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$