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Neko [114]
3 years ago
5

What is the molality of a solution of Fe(ClO3), in water that freezes at -2.72°C?

Chemistry
1 answer:
Anon25 [30]3 years ago
3 0

Answer:

0.366m = Molality of the solution

Explanation:

To solve this question we must know the addition of a solute produce decreasing in freezing point regard to the pure solvent. The equation is:

ΔT = m*Kf*i

<em>Where ΔT is change in freezing point </em>

(As freezing point of water is 0°C, the ΔT is 2.72°C)

<em>Kf is freezing point depression constant = 1.86°C/m for water</em>

<em>i is Van't Hoff factor. The number of ions produced when 1 mole of the salt is dissolved = 4 ions for Fe(ClO₃)₃, Fe³⁺ and 3 ClO₃⁻ ions</em>

<em>m is molality of the solution.</em>

<em />

Replacing:

2.72°C = m*1.86°C/m*4

<h3>0.366m = Molality of the solution</h3>

<em />

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Given that w for water is 2. 4×10−14 m^2 at 37 °C. Calculate the ph of a neutral aqueous solution at 37 °c, which is the normal
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The pH of a neutral aqueous solution at 37°C is 6.8.

<h3>What is Kw? </h3>

Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:

Kw = [OH⁻] . [H₃O⁺]

Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.

<h3>Calculation</h3>

Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:

Kw =[OH⁻] . [H₃O⁺] = x²

x² = 2.4 × 10⁻¹⁴ M²

x = 1.5919 × 10⁻⁷ M

Hence, the concentration of OH and H₃O⁺ (x) = [H₃O⁺] = [OH⁻] = 1.5919×10⁻⁷ M

pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)

pH = 6.8

Thus, we find that the pH of a neutral aqueous solution at 37 °c (which is the normal human body temperature) is 6.8.

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brainly.com/question/9529394

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