Visit the website 123iitjee.manishverma.site for latest posts, courses, admission & more.

### 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

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

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

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

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

$\therefore x = 2$

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$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, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$