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New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Solubility product of A? X = 4S? ³
Where S? is the solubility of salt A? X.
Solubility product of MX = S? ²
Where S? is the solubility of MX.
Given 4S? ³ = 4 * 10? ¹² ⇒ S? = 10? M
Given S? ² = 4 * 10? ¹² ⇒ S? = 2 * 10? M
So, S? / S? = 10? / (2 * 10? ) = 50

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

m=2 molal
ΔT? = 100.52 – 100 = 0.52°C
Using, ΔT? = iK? m
0.52 = I * 0.52 * 2
i = 0.5
Now using, α = (1-i) / (1-1/n)
Where, n=2 (dimerisation)
α = (1-0.5) / (1-0.5) = 1
So, percentage association = 100%

New answer posted

a month ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Please consider the following Image

 

New answer posted

a month ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

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.

New answer posted

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Molality = (mole of solute * 1000) / wt of solvent (gm)
100 = (n_solute * 1000) / [ (1 - n_solute) * 18]
(1 - n_solute) / n_solute = 1000 / (100 * 18) = 10/18
18 (1 - n_solute) = 10 n_solute
18 - 18 n_solute = 10 n_solute
18 = 28 n_solute
n_solute = 18 / 28? 0.6428 = 64.28 * 10? ²
Ans = 64 (Rounded off)

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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.

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Π = iCRT = I (n/V)RT
Πfinal = (Π? V? + Π? V? )/ (V? + V? )
Πfinal = [ (0.1 * 1) + (0.2 * 2)]/3
= (0.1 + 0.4)/3 = 0.5/3 = (500/3) * 10? ³ atm
so X = 167

New answer posted

a month ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Equivalent of solute = 0.1 * 0.1
Mole of solute (Na? CO? ·xH? O) = [0.1 * 0.1]/2
Mass of Na? CO? ·xH? O = ( [0.1 * 0.1]/2) * [106 + 18x] = 1.43
⇒ 106 + 18x = 286
18x = 180
x = 10

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