Chemistry Solutions

W = 4.5 g

M.W = 90

V = 250 ml

Using M = $\frac{W}{M.W*V}*1000$  

$M=\frac{4.5}{90*250}*1000=0.2$       

M = 2 * 10^-1 M

So, x = 2

$\Delta T_f=0.5°C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

$\Delta T_f=i{k}_f.m$           

$0.5=i*1.86*\frac{9.45}{94.5*500}*1000$          

i = 1.344

Now, using $\alpha =\frac{i-1}{n-1}$  

n for ClCH₂COOH = 2

 a =   $\frac{1.344-1}{2-1}=0.344$

Using

  $K_a=\frac{C{\alpha }^2}{1-\alpha }$           

$K_a=\frac{0.2*{\left(0.344\right)}^3}{0.66}=36*10^{-3}$           

           So; x = 36

i = total no. of particles after dissociation/association / total no. of particle before dissociation/association
HA? H? + A? (Dissociation)
0.5a
2HA → (HA)? (Association)
0.5a/2
i = (a + 0.25a)/a = 125 x 10? ²
Ans. = 125

$\pi =i*c*R*T=\frac{2*0.83*300}{60*10^3}=0.00415bar$

= 415 Pa

Using Raoult's Law, P_Total = P°_A·X_A + P°_B·X_B = (21 kPa * 1/3) + (18 kPa * 2/3) = 7 + 12 = 19 kPa.
Answer: 19 kPa

This is a limiting reactant problem. The initial moles are 5.25 mmol of Pb (NO? )? and 2.4 mmol of Cr? (SO? )? Based on the 3:1 stoichiometric ratio, Pb (NO? )? is the limiting reactant. The moles of PbSO? formed are equal to the initial moles of the limiting reactant, which is 5.25 mmol or 5.25 * 10? ³ moles. The question asks for the answer in units of 10? moles.

Using the Ideal Gas Law, PV = nRT:

P = 1 bar

V = 20 mL = 0.020 L

R = 0.083 L·bar·mol? ¹·K? ¹

T = 273 K

n = PV / RT = (1 * 0.020) / (0.0831 * 273) = 8.8 * 10? mol of Cl?

Number of Cl? molecules (N) = n * N_A = (8.8 * 10? ) * (6.022 * 10²³) = 5.3 * 10²? molecules.

Number of Cl atoms = 2 * (5.3 * 10²? ) = 1.06 * 10²¹.

The answer, rounded off to the nearest integer for the power of 10²¹, is 1.

There are multiple factors that make the carbonyl group a strong ligand. Check the list below for the reasons.

• Unlike other alkyl ligands, it is an unsaturated compound.
• Due to its unsaturated nature, it has difficulty donating? electron density.
• It has a tendency to accept? (Pie) antibonding electrons.
• CO ligand acts as Lewis acid and donates a lone pair of electrons to form a metal-carbon bond.
• The? -acidic nature of CO gives a strong field and greater d-orbital splitting.

Relative lowering in vapour pressure depends on no. of mole of solute greater the no. of mole of solute greater in RLVP and smaller will be vapour pressure. So order of vapour pressure is B > C > A.