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ivanzaharov [21]
2 years ago
11

52.1 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 38.5 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)

2 (aq) 2 NaCl (aq) --> PbCl2 (s) 2 NaNO3 (aq) The concentration of Pb2 ion in the solution is _____ M after the reaction is complete.
Chemistry
1 answer:
Elan Coil [88]2 years ago
6 0

Answer:

0.0585 M

Explanation:

  • Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)

First we <u>calculate the inital number of moles of each reagent</u>, using the <em>given volumes and concentrations</em>:

  • 0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂
  • 0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl

Then we <u>calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl</u>, using the <em>stoichiometric coefficients of the reaction</em>:

  • 16.0 mmol NaCl * \frac{1mmolPb(NO_3)_2}{2mmolNaCl} = 8.00 mmol Pb(NO₃)₂

Now we <u>calculate the remaining number of Pb(NO₃)₂ moles after the reaction</u>:

  • 13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂

Finally we <em>divide the number of moles by the final volume</em> to <u>calculate the concentration</u>:

  • 5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M
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Answer:

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Explanation:

As concentrations of two aqueous solutions are same therefore we can write:

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i = total number of ions generated from dissolution of one molecule of a substance (for strong electrolyte).

Here both Na_{2}CO_{3} and LiCl are strong electrolytes.

So, i(Na_{2}CO_{3})=3 and i(LiCl)=2

Hence, lowering of vapor pressure, elevation in boiling point and osmotic pressure will be higher for Na_{2}CO_{3} solution.

Therefore the Na_{2}CO_{3} solution has a higher osmotic pressure and higher boiling point than LiCl solution.

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