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Bezzdna [24]
3 years ago
8

Which of the following steps correctly describes the preparation of 0.5 M solution of FeCl3?

Chemistry
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

Option A is correct

Explanation:

Data given:

Steps for preparation of 0.5 M solution = ?

Solution:

Molarity: This the term used for the concentration of the solution. It is the amount of solute in moles dissolve in 1 Liter of solution.

Formula used for Molarity

         Molarity = number of moles of solute / 1 L of solution

From the above definition and formula it is clear that we have to know the following things.

  • no. of moles of f the solute
  • we also should know that of what concentration we are preparing solution in terms of molarity
  • The total amount of should be in liter of solution not solvent separately

Now:

As we couldn't weight mole on balance so have to convert moles to grams by using molar mass or mass in grams formula

As

   1 mole = mass in grams / molar mass

so to convert mole to mass rearrange the above equation

  mass in grams = 1 mole x molar mass

So,

1 mole of any substance equal to its molar mass

Now, to prepare 0.5 M solution of FeCl₃

  • First we find molar mass and weigh it

So,

1 mole of FeCl₃ = molar mass of FeCl₃

  • for 0.5 M

Multiply the molar mass with 0.5

  • To make 1 L of solution

As we know that molarity is the amount of solution in liter of solution not solvent, so we add the calculated amount of solute in volumetric flask and then add water till the 1 L mark. that is amount of solute plus solvent combined form 1 L of Solution.

So, Option A is correct: Find the molar mass of FeCl3 and multiply by 0.5 to find the amount. Add this amount in a volumetric flask and add water till the 1 L mark

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Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

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2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

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). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50
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Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C =  2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568  g C

Mass of H = 0.6955 g H₂O  × (2.016 H/18.02 g H₂O)  = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

=====

2. Convert these masses to moles.

Moles  C = 0.5568  × 1/12.01  = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654   × 1/16.00 = 0.054 09

=====

3. Find the molar ratios.

Moles  C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36   = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

======

4. Multiply the ratios by a number to make them close to integers

C  = 1        × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6  = 7.001

=====

5. Round the ratios to integers

C:H:O =6:10:7

=====

6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

=======

7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

=====

8. Divide the molecular mass by the empirical formula mass.  

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

=====

9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

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