Chemistry

KCl solution has molality (m) = 3.3, [Means 3.3 mol of KCl dissolved in 1 kg of solved]

Total mass of solution = mass of solute + mass of solvent

= 3.3 * 74.5 + 1000 gm

= 1245.85 gm

Volume of solution = $\frac{Mass}{density}=\frac{1245.85}{1.2}=1038.20ml$

$M=\frac{moles*1000}{V\left(ml\right)}=\frac{3.3*1000}{1038.2}=3.17M$          

Ans. = 3 (the nearest integer)

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

If Rate = $K{\left[NO\right]}^a{\left[H_2\right]}^b$  

Case I => $7*10^{-9}=K{\left(8*10^{-5}\right)}^a{\left(8*10^{-5}\right)}^b$

Case II =>   $2.1*10^{-8}=k{\left(24*10^{-5}\right)}^a{\left(8*10^{-5}\right)}^b$

Therefore, $\frac{7\text{?}*10^{-1}}{2.1}={\left(\frac{1}{3}\right)}^a$

or ${\left(\frac{1}{3}\right)}^1={\left(\frac{1}{3}\right)}^a\therefore a=1$           

Ans. = 1

Molecular formula of Mohr' salt, FeSO₄. (NH₄)₂SO₄.6H₂O.

Molecular formula of potash alum = K₂SO₄.Al₂ (SO₄)₃.24H₂O. Overall molecular formula of potash alum = KAl (SO₄)₂. 12H₂O. Ratio of water molecules in Mohr' salt and Potash alum

$=\frac{6}{12}=\frac{1}{2}=0.5=5*10^{-1}$       

Ans. = 5

All the solution have higher ion production with respect to 0.1 M C₂H₅OH i.e. 0.1 M Ba₃ (PO₄)₂, 0.1 M Na₂SO₄, 0.1 M KCl and 0.1 M Li₃PO₄. Hence all have lowered freezing point than 0.1 M C₂H₅OH (which is non-ionisable in aqueous medium).

Ans. = 4

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

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

$\therefore$ Lattice energy = 717.8

$\underset{\left(acid\right)}{H_{2}S}+\underset{Base}{H_{2}O}?H_3{O}^{+}+H{S}^{-}$

$\underset{\left(acid\right)}{H_{2}O}+\underset{\left(Base\right)}{N{H}_3}?N{H}_{4}OH$

With H₂S water acts like base and with NH₃ it acts like acid.

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

Elemente- gain enthalpy

F-333

Cl-349

Te-190

Po-174