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Alex
3 years ago
9

KCl is an ionic salt with a formula weight of 74.54 g/mole. A 1.417 M (mole/liter) KCl solution has a density of 1064.5 grams/li

ter at 20.0 OC. Use the Molarity and the formula weight to find the mass (g) of salt in one liter. Then, find the mass (g and kg) of water in one liter of the solution as the difference between the solution mass and the KCl mass. Include the conversion factor. Determine the molality (m), or moles of salt per kg of water, by dividing the moles of salt in one liter by the kg of water in one liter. Show all units and use proper significant figures. (
Chemistry
1 answer:
xenn [34]3 years ago
3 0

Answer:

Mass of the salt:  105.6g of KCl.

Mass water: 958.9g of water.

Molality: 1.478m.

Explanation:

<em>Mass of the salt:</em>

In 1L, there are 1.417 moles. In grams:

1.417 moles KCl * (74.54g / mol) = 105.6g of KCl

<em>Mass of the water:</em>

We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:

1L * (1064.5g / L) = 1064.5g - Mass solution.

Mass water = 1064.5g - 105.6g = 958.9g of water

<em>Molality:</em>

Moles KCl = 1.417 moles KCl.

kg Water = 958.9g = 0.9589kg.

Molality = 1.417mol / 0.9589kg = 1.478m

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Due to the fact that ice is less thick than liquid water, the phase diagram of water has an odd melting point that drops with pressure. Carbon dioxide cannot exist as a liquid at atmospheric pressure, according to the phase diagram of the gas. Thus, gaseous carbon dioxide directly sublimes from solid carbon dioxide.

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1 year ago
100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper
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Answer:

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃

               = (M)(Volume of the solution (L))

               = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn

           = Qrxn/(no. of moles of AgNO₃)

            = (1170.4 J)/(0.03 mol)

            = 39013.33 J/mol

           = 39.013 kJ/mol.

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3 years ago
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