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torisob [31]
3 years ago
12

a solution of unknown molecular substance is prepared by dissolving 0.50g of the unknown in 8.0g of benzene. the solution freeze

s at 3.9. determine the molar mass of the unknown
Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.

Explanation:

Mass of the unknown substance = 0.50 g

Freezing point of the solution = 3.9 °C

Freezing point of pure benzene = 5.5 °C

Freezing point dissociation constant Kf = 5.12°C/m

First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.

Change in temperature = 5.5 -3.9 = 1.6 °C

Next is to calculate the number of moles or molarity of the compound that dissolved.

Using the formula:

Δt = i Kf m

Assume i = 1

So,

1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene

x = 1.6 * 0.008 / 5.12

x = 0.0128 / 5.12

x = 0.0025 moles.

Next is to calculate the molar mass using the formula, molarity = mass / molar mass

Molar mass = mass / molarity

Molar mass = 0.50 g /0.0025

Molar mass = 200 g/mol

Hence, the molar mass of the unknown compound is 200 g/mol

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What is the energy needed to raise 2.004g of water from 89.0°C to 139.0°C?
gulaghasi [49]

Answer:

The answer to your question is E = 419.435 J

Explanation:

Data

Mass = m = 2.004 g

Temperature 1 = T1 = 89°C

Temperature 2 = T2 = 139°C

Specific heat  of water = Cp = 4.186 J/g°C

Energy = E = ?

Formula

E = mCp(T2 - T1)

Substitution

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Simplification and result

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6 0
3 years ago
Question 12 (5 points) During an experiment a student traps 0.1941 g of an unknown gas in a 175.6 mL container at 298.1 K and a
Tanzania [10]

Answer:

V₂ = 155.2 mL

Explanation:

Given data:

Initial volume =175.6 mL

Initial pressure =0.9648 atm

Initial temperature = 298.1 K

Final temperature = 273 K

Final volume =?

Final pressure = 1 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂  

V₂ = P₁V₁T₂  /T₁P₂

V₂ = 0.9648 atm × 175.6 mL × 273 k / 298.1 K× 1 atm

V₂ = 46251.4 atm. k. mL/ 298.1 K× 1 atm

V₂ = 155.2 mL

8 0
3 years ago
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