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spayn [35]
3 years ago
9

How many grams are in 3.93 x 10^24 molecules of CCl4?

Chemistry
1 answer:
Lorico [155]3 years ago
8 0
<h3>Answer:</h3>

1000 g CCl₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.93 × 10²⁴ molecules CCl₄

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.93 \cdot 10^{24} \ molecules \ CCl_4(\frac{1 \ mol CCl_4}{6.022 \cdot 10^{23} \ molecules \ CCl_4})(\frac{153.81 \ g \ CCl_4}{1 \ mol \ CCl_4})
  2. Multiply:                                                                                                             \displaystyle 1003.77 \ g \ CCl_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1003.77 g CCl₄ ≈ 1000 g CCl₄

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Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
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<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

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i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

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M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
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