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3 years ago

Assuming that seawater is a 3.50 mass % aqueous solution of NaCl, what is the molality of seawater?

Chemistry

1 answer:

Fantom [35]3 years ago

7 0

Hey there!

Consider 100 g of solution:

Mass of NaCl = 3.50% of mass of seawater

( 3.50 / 100 ) * 100 => 3.50 g

Number of moles as shown below:

Molar mass NaCl = 58.44 g/mol

n = Mass / molar mass

n = 3.50 / 58.44 => 0.059 moles of NaCl

Mass of sweater:

Mass of solution - Mass of NaCl

100 - 3.50 = 96.5 g

96.5 g in Kg :

96.5 / 1000 => 0.0965 Kg

Therefore ,calculate molality by using the following formula:

molality = number of moles of solute / mass of solution

molality = 0.059 / 0.0965

molality = 0.61 m

Hope That helps!

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1 year ago

Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in 3.72L vassel at a total pre

Julli [10]

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moles = 0.093 moles

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In this case, we know that this reaction is taking plave in a vessel that has a 730 torr of total pressure.

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The pressure of oxygen is:

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PV = nRT

With this expression, we will calculate the moles of O2, and then, the moles of KClO3:

n = PV/RT

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