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Zolol [24]
4 years ago
12

A solution is prepared by dissolving 10.0 g NaCl in 600 g of water. What is its molality? (Note: the molar masses are NaCl = 58.

5 g/mol; H2O = 18.0 g/mol)
A.
0.892m NaCl
B.
0.285m NaCl
C.
5.13m NaCl
D.
1.81m NaCl
Chemistry
1 answer:
xxMikexx [17]4 years ago
3 0

Answer:

Option B. 0.285m NaCl

Explanation:

To find the molality, first let find the number of mole of NaCl in the solution. This is illustrated below

Molar Mass of NaCl = 58.5 g/mol

Mass of NaCl = 10g

Number of mole of NaCl = Mass of NaCl / molar Mass of NaCl

Number of mole of NaCl = 10/58.5 = 0.1709mol

Mass of water = 600g = 600/1000 = 0.6kg

Molality = mole of solute / mass of solvent

Molality = 0.1709mol/0.6kg

Molality = 0.285mol/Kg

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General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
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<u>Organic</u>

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<u>Stoichiometry</u>

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

Avogadro's Number

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Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                   \displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O

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