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prohojiy [21]
3 years ago
11

Calculate the molarity of the final solution of Erythrocin B (MW 879.9 g/mol) if a .5028g sample is dissolved in water diluted t

o 100 mL in a volumetric flask (Solution 1), and then a 5mL aliquot of the stock solution is diluted to 100mL using volumetric equipment. (solution 2)
Chemistry
1 answer:
Mashcka [7]3 years ago
7 0

Answer:

2.857 × 10⁻⁴ M

Explanation:

First, we will calculate the molarity of the solution 1.

M = mass of solute / molar mass of solute × liters of solution

M = 0.5028 g / 879.9 g/mol × 0.100 L

M = 5.714 × 10⁻³ M

We can find the concentration of the final solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 5.714 × 10⁻³ M × 5 mL / 100 mL

C₂ = 2.857 × 10⁻⁴ M

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What mass of gold is produced when 24.4 a of current are passed through a gold solution for 46.0 min ?
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To solve this problem, we make use of the Faraday’s law for Electrolysis. The formula is given as:

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The gram equivalent weight is calculated by dividing the molar mass with the amount of charge produced per atom. Gold has charge of 3+ therefore the gram equivalent weight is:

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Solving for the mass m:

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A sample of pure lithium carbonate contains 18.8% lithium by mass. What is the % lithium by mass in a sample of pure lithium car
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Answer:

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So, % lithium by mass in second sample = (\frac{\frac{18.8}{100}\times 2x}{2x})\times 100 % = 18.8%

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Hope this helps!

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