Class 11th

Total number = 20 + 8 + 24 = 52

$x^2+y^2+{\left(x-1\right)}^2+y^2+x^2+{\left(y-1\right)}^2+{\left(x-1\right)}^2+{\left(y-1\right)}^2=18$

$\Rightarrow x^2+y^2-x-y-\frac{7}{2}=0$

$r^2=\frac{1}{4}+\frac{1}{4}+\frac{7}{2}=4$

$LHS={\displaystyle \sum _{r=1}^{15}r\left(r!\right)}$

$=\sum \left(r+1-1\right)r!={\displaystyle \sum _{r=1}^{15}\left(r+1\right)!-r!}$

$=\left(2!-1!\right)+\left(3!-2!\right)+....+\left(16!-15!\right)$

$\begin{array}{l}=16!-1\\ {=}^{16}{P}_{16}-1\end{array}$

Equation : 2x² – (6 + k)x + 4 + 3k = 0

D < 0 => 6  $-4\sqrt[]{2}<k<6+4\sqrt[]{2}$

Sum of integral values of K is 1+2+3+.+11 = 66

Side of square = a

Radius of circle = r

Given : 4a + 2pr = 36

S = a² + pr²

^{$=\left(\frac{{\pi }^2}{4}+\pi \right)r^2-9\pi r+81$}

^{$\frac{ds}{dr}=0\Rightarrow r=\frac{18}{\pi +4}$}

0.0504 M NH₄Cl of 5ml => millimole of $N{H}_4^{+}=0.0504*5$

0.0210 M NH₃ of 2ml => millimole of NH₃ = 0.0210 * 2

It is a basic buffer.

Total volume = 7ml

$Here,K_b=\frac{\left[O{H}^{-}\right]*\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$     

$\therefore 1.8*10^{-5}=\frac{\left[O{H}^{-}\right]*0.0504*5}{0.0210*2}$         

$\therefore \left[O{H}^{-}\right]=\frac{1.8*10^{-5}*0.0210*2}{0.0504*5}=0.3*10^{-5}M$  

$\therefore \left[O{H}^{-}\right]=3*10^{-6}M$

$\therefore x=3$   

Ans. = 3

$\Delta H_f^0=\left(+\Delta H_{sub}^0\right)+\left(+\Delta H_{IE}^0\right)+\left(+\frac{1}{2}\Delta H_{BDE}^0\right)+\left(\Delta H_{EA}^0\right)+\left(-\Delta H_{LE}^0\right)$

$\Delta H_{LE}^0=89.2+419+121.5-348.6+436.7kJmo{l}^{-1}$

$\therefore$ Lattice energy = 717.8