Ncert Solutions Chemistry Class 11th

Statement I is true because methyl orange indicator has pH range (3.2 to 4.2) which is suitable for buffer produced by strong acid and weak base.

Statement II is false because phenolphthalein has pH range (8.4 to 10.2) and only suitable for equivalence point above than 7, In this case pH increased by the addition of NaOH on acetic acid.

De-ionised water means dissolved minerals free water which is prepared by synthetic resin method. Where cation exchanged by H⁺ and anion exchanged by OH^- of resin.

$Ca{\left(OH\right)}_2+C{O}_2\to CaC{O}_3+H_{2}O$ (1st step of reaction)

$CaC{O}_3+H_{2}O+C{O}_2\to Ca{\left(HC{O}_3\right)}_2$ (2nd step of reaction)

In alkaline earth metals cationic size increases down the groups, hence thermal stability of carbonates increases down the group. Therefore both assertion and reason are correct and reason is correct explanation of assertion.

$P=\frac{nRT}{V}=\frac{\left(1+2\right)*8.3*400}{4.0*10^{?3}}=24.9*10^{5}Pa$

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