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uysha [10]
2 years ago
12

Match the solutions to the descriptions of the freezing points.a. One mole of the ionic compound Na3PO4 dissolved in 1000 g H2O

b. One mole of the ionic compound CuSO4 dissolved in 1000 g H2O c. One mole of the nonelectrolyte C6H12O6 dissolved in 1000 g H2O1. Highest freezing point 2. Intermediate freezing point 3. Lowest freezing point
Chemistry
1 answer:
rosijanka [135]2 years ago
5 0

Explanation:

Depression in Freezing point

= Kf × i × m

where m is molality , i is Van't Hoff factor, m = molality

Since molality and Kf remain the same

depression in freezing point is proportional to i

i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2

i=1 for C2h6O

i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)

So the freezing point depression is highest for MgCl2 and lowest for C2H6O

so freezing point of the solution = freezing point of pure solvent- freezing point depression

since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point

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1. The molar mass of the unknown gas obtained is 0.096 g/mol

2. The pressure of the oxygen gas in the tank is 1.524 atm

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>1. How to determine the molar mass of the gas </h3>
  • Rate of unknown gas (R₁) = 11.1 mins
  • Rate of H₂ (R₂) = 2.42 mins
  • Molar mass of H₂ (M₂) = 2.02 g/mol
  • Molar mass of unknown gas (M₁) =?

R₁/R₂ = √(M₂/M₁)

11.1 / 2.42 = √(2.02 / M₁)

Square both side

(11.1 / 2.42)² = 2.02 / M₁

Cross multiply

(11.1 / 2.42)² × M₁ = 2.02

Divide both side by (11.1 / 2.42)²

M₁ = 2.02 / (11.1 / 2.42)²

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<h3>2. How to determine the pressure of O₂</h3>

From the question given above, the following data were obtained:

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  • Mass of O₂ = 0.885 kg = 885 g
  • Molar mass of O₂ = 32 g/mol
  • Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
  • Temperature (T) = 21 °C = 21 + 273 = 294 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
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The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side by V

P = nRT / V

P = (27.65625 × 0.0821 × 294) / 438

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Learn more about Graham's law of diffusion:

brainly.com/question/14004529

Learn more about ideal gas equation:

brainly.com/question/4147359

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How many moles are present in 63.80L of oxygen gas, O2?
matrenka [14]

Answer:

2.85moles of oxygen gas

Explanation:

Given parameters:

Volume of oxygen gas  = 63.8L

Unknown:

Number of moles  = ?

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We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:

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 So;

       63.8L of oxygen gas will take up a volume of

              \frac{63.8}{22.4}   = 2.85moles of oxygen gas

4 0
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