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Answer : Option A) 0.5 M NaCl.

Explanation : A solution can be called as a stronger electrolyte which when dissolved in aqueous solution and produce ions in the solutions and is easily decomposed by electrolysis. Amongst the given options NaCl produces Na ions and Cl ions on dissolving with water rest all are not so readily soluble in water. As only salts which belong to Group I elements are soluble electrolytes.

6 0

3 years ago

If 4.27 g sucrose (c12h22o11) are dissolved in 15.2 g water, what is the boiling point of the resulting solution? kb for water =

NikAS [45]

Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
m(C₁₂H₂₂O₁₁) = 4.27 g.
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.

4 0

3 years ago

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The dissolution of 7.75 g of a substance in 825 g of benzene (C6H6) at 298 K raises the boiling point by 0.575 ◦C. Note that kf

Dennis_Churaev [7]

Answer:

a) The freezing point depression = 1.16 °C

b) Psolution = 101.2 torr Psolution/Psolvent = 101.2 / 103 = 0.983

c) osmotic pressure π = 4.85 atm

d) Molecular weight = 41.4 g/mol

Explanation:

Step1 : Data given

Mass of substance = 7.75 grams

Mass of benzene = 825 grams

Boiling point elevation = 0.575 °C

Kf = 5.12 C/m

Kb = 2.53 C/m

Density = 876.6 Kg /m³ = 0.8766 g/mL

Step 2 Calculate the molaltiy

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = 0.575 °C

⇒i = the van't Hoff factor = 1

⇒Kb = the boiling point constant = 2.53 °C/m

⇒with m = TO BE DETERMINED

0.575 = 2.53 * m

m = 0.227 molal

Step 3: Calculate the freezing point depression

ΔT = i*Kf*m

⇒ΔT = the freezing point depressio, = TO BE DETERMINED

⇒i = the van't Hoff factor = 1

⇒Kb = the freezing point constant = 5.12 °C/m

⇒with m = 0.227 molal

ΔT = 5.12 °C/m * 0.227 m

ΔT = 1.16 °C

Step 4: Calculate the molar mass of the solute

Molality = moles / mass benzene

0.227 molal = moles / 0.825 kg

Moles substance = 0.227 * 0.825

Moles substance = 0.187 moles

Molar mass = mass / moles

Molar mass substance = 7.75 grams / 0.187 moles

Molar mass substance = 41.4 g/mol

Step 5: Calculate the ratio of the vapor pressure above the solution to that of the pure solvent,

Moles benzene = 825 grams / 78.11 g/mol

Moles benzene = 10.56 moles

Mol fraction benzene = 10.56 moles /(10.56 + 0.187 )

Mol fraction benzene = 0.983

Psolution = 103 * 0.983 = 101.2 atm

Step 5: Calculate moles substance

Moles substance = 7.75 grams / 41.4 g/mol

Moles substance = 0.187 moles

Step7: Calculate volume

Volume = mass / density

Volume = 825 * 10^-3 / 876.6

Volume = 0.942 L

Step 7 Calculate osmotic pressure

π = i*M*R*T = i n/V * R*T

⇒π = the osmotic pressure = TO BE DETERMINED

⇒i = the van't Hoff factor = 1

⇒M = the molar concentration = n/V

⇒R = the gas constant = 0.08206 L*atm/mol*K

⇒T = the temperature = 298 K

π = 1 * (0.187 moles / 0.942L) * 0.08206 * 298 K

π = 4.85 atm

5 0

3 years ago