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lora16 [44]
2 years ago
5

6. A 25.0-mL sample of potassium chloride solution was found to have a mass of 25.225 g.

Chemistry
1 answer:
Morgarella [4.7K]2 years ago
4 0

Answer:

6. a. 5.53 (m/m) %; b. 0.7490M

7. 0.156M

Explanation:

6. In the solution of KCl, there are 1.396g of KCl in 25.225g of solution. As mass/mass percent is:

Mass solute / Mass solution * 100

The mass/mass percent of KCl is:

a. 1.396g KCl / 25.225g solution * 100

5.53 (m/m) %

b. Molarity is moles of solute per liters of solution:

<em>Moles KCl -Molar mass: 74.55g/mol-:</em>

1.396g KCl * (1mol / 74.55g) = 0.018726 moles KCl

<em>Volume in liters: 25mL = 0.025L</em>

Molarity:

0.018726 moles KCl / 0.025L = 0.7490M

7. 0.90% means 0.90g of NaCl in 100g of solution:

<em>Moles NaCl -Molar mass: 58.44g/mol-:</em>

0.90g NaCl * (1mol / 58.44g) = 0.0154 moles NaCl

<em>Volume in Liters:</em>

100g * (1mL / 1.01g) = 99mL = 0.099L

Molarity:

0.0154 moles NaCl / 0.099L =

0.156M

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Write the fraction of the mass of kcl produced from 1 g of k2c03​
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<h3>Further explanation</h3>

Given

1 g of K₂CO₃

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Mass of KCl

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The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

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