Solutions

AB2 is 10% dissociated in water to A²⁺ and B^-. The boiling point of a 10.0 molal aqueous solution of AB2 is_________ °C. (Round off to the nearest Integer).

[Given : Molal elevation constant of water Kb = 0.5 K kg mol^-1, boiling point of pure water = 100°C]

0 Follower 4 Views

Contributor-Level 9

m = 10 molal
K_b = 0.5 K kg mol? ¹
Using: ΔT_b = I K_b m
and α = (i - 1) / (n - 1)
n for AB? is 3; α = 0.1
0.1 = (i - 1) / (3 - 1) ⇒ I = 1.2
ΔT_b = 1.2 * 0.5 * 10 = 6 °C
So, boiling point of solution = 100 + 6 = 106 °C

0 0

New answer posted

3 months ago

A 6.50 molal solution of KOH (aq) has a density of 1.89 g cm^-3. The molarity of the solution is________ mol dm^-3. (Round off to the nearest Integer).

[Atomic masses : K = 39.0u, O = 16.0u, H= 1.0u]

0 Follower 2 Views

Contributor-Level 9

A 6.5 molal solution means 6.5 moles of KOH is in 1 kg (1000 g) of solvent (H? O).
Moles of solute, n_B = 6.5
Mass of solute, W_B = 6.5 * 56 = 364 g
Mass of solvent, W_A = 1000 g
Mass of solution = 1364 g
Volume of solution = 1364 / 1.89 mL
Now, molarity = [6.5 / (1364 / 1.89)] * 1000 M = 9 M

0 0

New answer posted

3 months ago

Two salts A2X and MX have the same value of solubility product of 4.0 x 10^-12. The ratio of their molar solubilities i.eS A2X/S MX = ____________.

(Round off to the Nearest Integer).

0 Follower 2 Views

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

0 0

New answer posted

3 months ago

At 363 K, the vapour pressure of A is 21 kPa and that of B is 18 kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is __________ kPa. (Round off to the nearest integer).

0 Follower 2 Views

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

0 0

New answer posted

3 months ago

A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52°C. The percentage association of A is ________. (Round off to the nearest integer).
[ Use: K? for water = 0.52 K kg mol?¹, Boiling point of water = 100°C]

0 Follower 2 Views

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%

0 0

New answer posted

3 months ago

2NO(g) + Cl?(g) ? 2NOCl(s)
This reaction was studied at -10°C and the following data was obtained

[NO]? and [Cl?]? are the initial concentrations and r? is the initial reaction rate. The overall order of the reaction is _____. (Round off to the nearest integer).

0 Follower 1 View

Contributor-Level 9

Please consider the following Image

0 0

New answer posted

4 months ago

The number of chlorine atoms in 20 mL of chlorine gas at STP is ______10²¹ (Round off to the nearest integer).
[Assume chlorine is an ideal gas at STP, R = 0.083 L bar mol⁻¹ K⁻¹, NA = 6.023 *10²³]

0 Follower 7 Views

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.

0 0

New answer posted

4 months ago

Solutions Class 11th

The mole fraction of a solute in a 100 molal aqueous solution is _______ *10^-2. (Round off to the nearest integer).
[Given : Atomic masses : H : 1.0u, O : 16.0u]

0 Follower 3 Views

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)

0 0

New answer posted

4 months ago

A soft drink was bottled with a partial pressure of CO? of 3 bar over the liquid at room temperature. The partial pressure of CO? over the solution approaches a value of 30bar when 44 g of CO? is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is * 10?¹ /
(First dissociation constant of H?CO? = 4.0 * 10?; log 2 = 0.3; density of the soft drink = 1 g mL?¹)

0 Follower 2 Views

Contributor-Level 10

0 0

New answer posted

4 months ago

A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A = 100 g mol⁻¹; B = 200 g mol⁻¹; C = 10,000 g mol⁻¹ ]

0 Follower 2 Views