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2 years ago

How many water molecules are found within the crystalline structure of one hydrate molecule?

Chemistry

1 answer:

den301095 [7]2 years ago

4 0

a. The Mass of water driven off = 0.15 g

b. Moles of anhydrate = 0.00257 moles

c. Moles of water driven off is 0.00833 moles

d. There are 3 moles of water within the crystalline structure of one molecule of the hydrated salt.

e. The molecular formula of the hydrated salt will be X.3H₂O

<h3>What is the mass of water driven off from the hydrated salt?</h3>

a. The mass of water driven off from the hydrated salt is:

Mass of water driven off = 0.5 g - 0.35 g

Mass of water driven off = 0.15 g

b. Molecular mass of salt = 136 g/mol

moles of anhydrate = 0.35/136

Moles of anhydrate = 0.00257 moles

c. Moles of water driven off = mass/molar mass

molar mass of water = 18 g/mol

Moles of water driven off = 0.15/18

Moles of water driven off = 0.00833 moles

d. Moles of water within the crystalline structure of one molecule of the hydrated salt is determined by converting to whole number mole ratio by dividing with the smallest ratio,

Salt to water ratio = 0.00257 /0.00257 : 0.00833/0.00257

Salt to water ratio = 1 : 3

Therefore, there are 3 moles of water within the crystalline structure of one molecule of the hydrated salt.

e. Assuming the anhydrous salt is X, the molecular formula of the hydrated salt will be X.3H₂O

Learn more about hydrated salts at: brainly.com/question/14447094

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liraira [26]

Answer:

The Henry's law constant for argon is $k=2.11*10^{-3}\frac{ M}{atm}$

Explanation:

Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.

C = k*P

where C is the solubility, P the partial pressure and k is the Henry constant.

So, being the concentration $C=\frac{ngas}{V}$

where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So $n=\frac{P*V}{R*T}$

In this case:

P=PAr= 1 atm
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R=0.082 $\frac{atm*L}{mol*K}$
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Then:

$n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}$

Solving:

n= 2.11 *10⁻³ moles

So: $C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M$

Using Henry's Law and being C=CAr and P =PAr:

2.11*10⁻³ M= k* 1 atm

Solving:

$k=\frac{2.11*10^{-3} M}{1 atm}$

You get:

$k=2.11*10^{-3}\frac{ M}{atm}$

The Henry's law constant for argon is $k=2.11*10^{-3}\frac{ M}{atm}$

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