Solutions
Get insights from 211 questions on Solutions, answered by students, alumni, and experts. You may also ask and answer any question you like about Solutions
- 211
Questions
- 0
Discussions
- 28
Active Users
- 0
Followers
Have a question about Solutions? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
Molality of sulphuric acid solution in which the mole fraction of water is 0.74 is:
Molality of sulphuric acid solution in which the mole fraction of water is 0.74 is:
? Solvent is H? O, which is in excess
So using m ( molality ) = (x? ×1000)/ (x? × (M? )? )
? x? = 0.74 (Mol? = 18 g)
x? = 1 – 0.74 = 0.26 ∴ m = (0.26 × 1000)/ (0.74 × 18) = 19.5
Upvote 
4.5g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is × 10-1. The value of x is_________ (Rounded off to the nearest integer)
4.5g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is × 10-1. The value of x is_________ (Rounded off to the nearest integer)
W = 4.5 g
M.W = 90
V = 250 ml
Using M =
M = 2 × 10-1 M
So, x = 2
Upvote 10.30 mg of O2 is dissolved in 1L of sea water of density 1.03 g/mL. The concentration of O2 in ppm is _______.
10.30 mg of O2 is dissolved in 1L of sea water of density 1.03 g/mL. The concentration of O2 in ppm is _______.
ppm of O? = (wt. of O? ) / (wt. of H? O) × 10?
= (10.3 mg) / (1.03 × 10? mg) × 10?
= 10ppm
Upvote In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van’t Hoff factor of the acid is _____ ×10⁻². (Round off to the Nearest Integer).
In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van’t Hoff factor of the acid is _____ ×10⁻². (Round off to the Nearest Integer).
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
Upvote In one molal solution that contains 0.5 mole of a solute, there is:
In one molal solution that contains 0.5 mole of a solute, there is:
Molality = moles of solute / mass of solvent (kgs) ⇒ 1 = 0.5 / W (kgs)
Wsolvent (kgs) = 0.5 = 500 g
Upvote 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]
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]
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
Upvote 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]
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]
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
Upvote 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).
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).
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
Upvote 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).
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).
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
Upvote 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]
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]
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%
Upvote