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Taya2010 [7]
3 years ago
11

When 16.9 g KOH is dissolved in 90.8 g of water in a coffee-cup calorimeter, the temperature rises from 18.5 °C to 34.27 °C. Wha

t is the enthalpy change per gram (in J/g) of KOH dissolved in the water? Assume that the solution has a specific heat capacity of 4.18 J/g×K. Water has a density of 1.00 g/ml. Be sure to enter the correct sign (+/-). Enter to 1 decimal place.
Chemistry
1 answer:
juin [17]3 years ago
4 0

Answer:

-431.5 J/g

Explanation:

Mass of solution = Mass of solute + mass of solvent

Solute is KOH while solvent is water.

Mass of KOH = 16.9 g

Mass of water = 90.8 g

Mass of solution = 16.9 + 90.8

                          = 107.7 g

Change in temperature (Δt) = 34.27 - 18.5

                                           = 16.2 °C

Heat required to raise the temperature of water is released by dissolving KOH.

Therefore,

Heat released by KOH = m × s× Δt

                                    = 107.7 × 4.18 × 16.2

                                    = 7293 J

Heat released by per g KOH = 7293 J/16.9 g

                                             = 431.5 J/g

As heat is released therefore, enthalpy change would be negative.

Enthalpy change of KOH = -431.5 J/g

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A student is given 2.19 g of an unknown acid, which can be either oxalic acid, H2C2O4, or citric acid, H3C6H5O7. To determine wh
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Answer:

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Explanation:

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Mass of the unknown acid = 2.19 gram

Titrating with 0.560 M of NaOH

The equivalence point is reached when 61.0 mL are added

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Molar mass of citric acid = 192.12 g/mol

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H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

The reaction between citric acid and NaOH is:

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<u>Step 3:</u> Calculate the number of moles of the acid

Moles = mass / Molar mass

In case of oxalic acid: 2.19 grams / 90.03 g/mol = 0.0243 moles

In case of citric acid: 2.19 grams /192.12 g/mol = 0.0114 moles

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For each mole of citric acid 3 moles of NaOH is required, for 0.0114 mol citric acid 0.0114 * 3= 0.0342 mol NaOH is required. This is the number of moles NaOH used for the titration.

The unknown acid is citric acid. There is 0.0342 moles of NaOH consumed.

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