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2 years ago

An aqueous solution contains 32.7% KCl (wt/wt%). what is the mole fraction of KCl in the solution

Chemistry

1 answer:

inna [77]2 years ago

5 0

The mole fraction of KCl in the solution is 0.1051

calculation

mole fraction of KCl in solution = moles of KCl / total number of moles(moles of KCl +moles of H2O)

moles=mass/molar mass

mass of KCl=32.7g

molar mass of KCl= 39 +35.5

moles of KCl is therefore= 32.7g/74.5 g/mol=0.439 moles

find the moles of H2O= mass of H2O/molar mass

mass of H2O=100-32.7=67.3g

molar mass of H2O=( 1 x2) +16=18 g/mol

moles = 67.3/18 =3.739 moles

total moles=3.739+0.439=4.178 moles

mole fraction is therefore=0.439/4.178=0.1051

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Answer :

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Explanation :

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Ionic compound are usually formed when a metal reacts with a non-metal.

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The covalent compound are usually formed when two non-metals react.

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As we know that the molecular compounds are the covalent compounds.

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This is an molecular compound because it is made up of sulfur and chlorine non-metal elements.

(d) Ca(NO₃)₂

This is an ionic compound because it is made up of calcium metal and polyatomic ion $(NO_3^-)$ non-metal elements.

(e) FeCl₃

This is an ionic compound because it is made up of iron metal and chlorine non-metal elements.

(f) LaP

This is an ionic compound because it is made up of lanthanum metal and phosphorous non-metal elements.

(g) CoCO₃

This is an ionic compound because it is made up of cobalt metal and polyatomic ion $(CO_3^{2-})$ non-metal elements.

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This is an ionic compound because it is made up of ammonium ion and sulfate ion.

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2 years ago

What happens if an atom loses a neutron?Its mass is lowered, but it is still the same element. Its mass stays the same, and it i

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2 years ago

Read 2 more answers

A single alkyl bromide reactant theoretically yields either of the given products, depending on the reaction conditions. Draw th

Ierofanga [76]

Answer:

$H_{3}C$ -- $H_{3}C$ -- $CH_{3}$ --Br-- $CH_{3}$

Explanation:

The steps involved in predicting the structure of the alkyl bromide compound are outlined below.

1) An examination of the product shows that the product could only be formed by a substitution reaction.

2) The structure of the alkyl bromide compound can be then predicted by replacing the methoxide group in the product after the substitution of bromine atom. This is because the methoxide ion acts as a strong nucleophile.

Therefore, by consideration the reaction mechanisms of reactions 1 and 2, it can be predicted that the structure of the alkyl bromide compound is $H_{3}C$ -- $H_{3}C$ -- $CH_{3}$ --Br-- $CH_{3}$ . A pictorial diagram of the alkyl bromide compound is also attached.