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Marizza181 [45]
3 years ago
7

To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?

Chemistry
1 answer:
Blizzard [7]3 years ago
8 0

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

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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

M₁ = 0.096 g/mol

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

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

  • Volume (V) = 438 L
  • 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
  • Pressure (P) =?

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

P = 1.524 atm

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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