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zepelin [54]
3 years ago
8

Calculate the mole fraction of each component in a solution with 6.87 g of sodium chloride (NaCl) dissolved in 65.2 g of water.

Work must be shown in order to earn credit.
Chemistry
1 answer:
zheka24 [161]3 years ago
7 0

Answer:

mole fraction of NaCl = 0.03145.

mole fraction of water = 0.9686.

Explanation:

  • Mole fraction is an expression of the concentration of a solution or mixture.
  • It is equal to the moles of one component divided by the total moles in the solution or mixture.
  • The summation of mole fraction of all mixture components = 1.

mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles).

<em>no. of moles of NaCl = mass/molar mass </em>= (6.87 g)/(58.44 g/mol) = 0.1176 mol.

<em>no. of moles of water = mass/molar mass</em> = (65.2 g)/(18.0 g/mol) = <em>3.622 mol.</em>

<em></em>

∴ mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles) = (0.1176 mol)/(0.1176 mol + 3.622 mol) = 0.03145.

<em>∵ mole fraction of NaCl + mole fraction of water = 1.0.</em>

∴ mole fraction of water = 1.0 - mole fraction of NaCl = 1.0 - 0.03145 = 0.9686.

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An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4, the concentration of hydronium ion in this solution (m) is 2.234 × 10⁻¹² M.

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CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Then, we use the ICE analysis which stands for Initial-Change-Equilibrium.

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Initial                    0.05          -                 0         0

Change                 -x                              +x       +x

----------------------------------------------------------------------------

Equilibrium         0.05-x                           x          x

Then, we use the equation for the equilibrium constant of basicity.

Kb = [CH₃NH₃][OH⁻]/[CH₃NH₂] = 4.4×10⁻⁴

4.4×10⁻⁴ = [x][x]/[0.05-x]

[x] = 4.4756×10⁻³

Here, variable x denotes the number of moles of the substance which is involved in the reaction. Since the equilibrium amount of OH⁻ is equal to x, then the concentration of OH⁻ is also 4.4756×10⁻³. Thus,

pOH = -log[OH⁻]

pOH = -log[4.4756×10⁻³] = 2.35

The relationship between pOH and pH is that pH + pOH = 14. Thus,

pH = 14 - 2.35 = 11.65

pH = -log[H⁺]

11.65 = -log[H⁺]

[H⁺] = 2.234 × 10⁻¹² M

Thus, we find the concentration of solution is 2.234 × 10⁻¹² M.

Learn more about aqueous solution: brainly.com/question/11097800

#SPJ4

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