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igomit [66]
3 years ago
7

How many grams of potassium chloride (KCl) can be dissolved in 68.0g of water at 40°C? Please help.

Chemistry
1 answer:
Molodets [167]3 years ago
5 0

Answer:

The problem provides you with the solubility of potassium chloride,

KCl

, in water at

20

∘

C

, which is said to be equal to

34 g / 100 g H

2

O

.

This means that at

20

∘

C

, a saturated solution of potassium chloride will contain

34 g

of dissolved salt for every

100 g

of water.

As you know, a saturated solution is a solution that holds the maximum amount of dissolved salt. Adding more solid to a saturated solution will cause the solid to remain undissolved.

In your case, you can create a saturated solution of potassium chloride by dissolving

34 g

of salt in

100 g

of water at

20

∘

C

.

Now, your goal here is to figure out how much potassium chloride can be dissolved in

300 g

of water at this temperature. To do that, use the given solubility as a conversion factor to take you from grams of salt to grams of water

300

g H

2

O

⋅

34 g KCl

100

g H

2

O

=

102 g KCl

You should round this off to one sig fig, since that is how many sig figs you have for the mass of water

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

mass of KCl

=

100 g

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−

Explanation:

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Here is the full question

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction: NH4NO3(s)→NH+4(aq)+NO−3(aq) In order to measure the enthalpy change for this reaction, 1.25 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 ∘C and the final temperature (after the solid dissolves) is 21.9 ∘C. Part A Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.0g/mL as the density of the solution and 4.18J/g⋅∘C as the specific heat capacity.) Express your answer to two significant figures and include the appropriate units. ΔHrxn =   ??? kJ/mol

Answer:

26 kJ / mol

Explanation:

Given that;

The mass of NH₄NO₃ = 1.25 g

Number of moles of NH₄NO₃ = Mass of NH₄NO₃ / Molar Mass of NH₄NO₃

Number of moles of NH₄NO₃= 1.25 g / 80.043 g/mol

Number of moles of NH₄NO₃= 0.016 mol

Volume of solution = 25.0 mL

Density of Solution = 1.0g/mL

Since; density = \frac {mass} {volume}

Mass of Solution = Density x Volume

= 1.0g/mL × 25.0mL

= 25 g

Heat Generated (Q)  = mc \delta T

Q= 25g × 4.18 J/g°C x (25.8°C - 21.9°C)

Q =  407.55 J

Q = 407 × 10 ⁻³ kJ

Q = 0.40755 kJ

Δ H_{rxn} = \frac{Heat generated(Q)}{number of molesof NH_4NO_3}

=  \frac {0.40755 kJ}{ 0.016 mol}

= 25.47 kJ/ mol

~ 26 kJ / mol

Therefore, the change in enthalpy for the reaction in kilojoules per mole = 26 kJ / mol

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Answer:

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Explanation:

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