Ncert Solutions Chemistry Class 11th

Acid = M (OH)₂ ® Salt + H₂O

M.E of Acid = me of Base

$\therefore 10*\left(0.1*nf\right)=\left(0.05*2\right)*30$

$\therefore n_f=3$

Thus basicity of acid = 3

$T_1=27°C=300K$  T₂ = 45° C = 45 + 273 = 318 K

P₁ = 30 atm        P₂ =?

V is constant due to rigid tank

$\therefore \frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{30}{300}=\frac{P_2}{318}\therefore P_2=\frac{1}{10}*318=31.8atm\approx 32atm$    

Total concentration of H+ after the mixing of HCl and H2SO4;

$\left[H^{+}\right]=\frac{\left(200*0.01\right)+\left(400*2*0.01\right)}{600}$

$\left[H^{+}\right]=\frac{1}{60}$

$\therefore pH=lo{g}_{10}\left(\frac{1}{60}\right)=1.78$

Mass of pure carbon in coal = 0.6 * 1000gm * $\frac{60}{100}=360gm$

Mass of carbon converted into CO = 6 $\frac{60}{100}*360=216gm\Rightarrow Mol=\frac{216}{12}=18$ mole of carbon

Mass of carbon converted into CO₂ = 360 – 216 = 144gm Mole = $\frac{144}{12}=12$ mole of carbon.

C (s) + O₂ (g) CO₂ (g) + 400KJ

Mole – 12 for CO₂ production, Hence total energy produced = 400 * 12 = 4800KJ

$C_{\left(s\right)}+\frac{1}{2}{O}_2\left(g\right)\underset{}{\overset{}{\to }}C{O}_{\left(g\right)}+100KJ$

Mole = 18 for CO production, Hence energy produced = 100 * 18 = 1800KJ

Total heat = 4800 + 1800 = 6600KJ

$4HNO{I}_{3\left(\mathcal{l}\right)}+3KC{l}_{\left(s\right)}\underset{}{\overset{}{\to }}C{l}_{2\left(g\right)}+NOC{l}_{\left(g\right)}+2H_{2}O\left(g\right)+3KN{O}_3\left(g\right)$

$\downarrow$

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$\therefore$ 1 mole of KNO₃ produced by $\frac{4}{3}$ moles of HNO₃

and $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4*110}{3*101}$ moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃ = mole * mol.wt = 145 * 63 = 91.48 $\approx$ 91.5gm

$\Delta H=3{\left(\Delta H_{comb}\right)}_{C_2{H}_2\left(g\right)}-{\left(\Delta H_{comb}\right)}_{C_6{H}_6\left(l\right)}$

$=-1300*3-\left(-3268\right)kJ/mole$

$=\left(-3900+3268\right)kJ/mole$

$=-632kJ/mole$

$C_{P,m}=C_{v,m}+R$

$C_{v,m}=20.785-8.314=12.471J{K}^{-1}mo{l}^{-1}$

$n=\frac{5000}{12.471*200}=\frac{25}{12.471}\approx 2$

$K{O}_2,N{O}_2,Cl{O}_2,NO$ are paramagnetic.

For an acid- base titration, Methly orange exist at end point as quinonoid form.