Ncert Solutions Chemistry Class 11th

Here, total meq of acetic acid = 50 * 0.1 = 5

And total meq of NaOH = 25 * 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$pH=pKa+lo{g}_{10}\frac{\left[S\right]}{\left[A\right]}$

$pH=4.76+lo{g}_{10}\frac{2.5}{2.5}=4.76=476*10^{-2}$  

Hence, x = 727 (the nearest integer)

Borax forms basic buffer solution in aqueous medium

Baking soda = NaHCO₃

Washing soda = Na₂CO₃. 10H₂O

Caustic soda = NaOH

HOCl produce in the stratospheric cloud, by the hydrolysis reaction of ClONO₂.

$ClON{O}_{2\left(g\right)}+H_2{O}_{\left(g\right)}\to HOC{l}_{\left(g\right)}+HN{O}_{3\left(vap\right)}$

Due to H- bond in water, it has high melting point and melting point of other hydrides of the group are depending upon the molecular weight.

Due to high crystallity Be has the highest M.P.

Be = 1560 K

Mg = 925 K

Ca = 1120 K

Sr = 1062 K

In boron family lower O.S (+1) is more stabilize down the group due to inert pair effect.

Wt of liquid = 135 – 40 = 95 gm

Volume of liquid =  $\frac{wt.}{density}=\frac{95}{0.95}=100ml$  

Hence volume of vessel = 100 ml = 0.1 lit from ideal gas equation,

$M=\frac{dRT}{P}=\frac{0.5}{0.1}*\frac{0.0821*250}{0.82}$  

$\therefore M=125g/mol$  

For expansion in vacuum, workdone, w = 0

For isothermal process,   $\Delta U=0$  

According to first law of thermodynamics,

$\begin{array}{l}\Delta U=q+w\\ q=0\end{array}$