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3 years ago

9

Whats the Boiling point of a Solution made by dissolving 31 g of NaCl of Water?

1 answer:

Anna11 [10]3 years ago

5 0

101° C is the answer.

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2 years ago

A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 295 mL of solution. How many moles o

Masja [62]

Answer:

(a) Moles of ammonium chloride = 0.243 moles

(b) $Molarity_{ammonium\ chloride}=0.824\ M$

(c) 60.68 mL

Explanation:

(a) Mass of ammonium chloride = 13.0 g

Molar mass of ammonium chloride = 53.491 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{13.0\ g}{53.491\ g/mol}$

<u>Moles of ammonium chloride = 0.243 moles</u>

(b) Moles of ammonium chloride = 0.243 moles

Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{ammonium\ chloride}=\frac{0.243}{0.295}$

$Molarity_{ammonium\ chloride}=0.824\ M$

(c) Moles of ammonium chloride = 0.0500 moles

Volume = ?

Molarity = 0.824 M

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$0.824\ M=\frac{0.0500}{Volume}$

<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>

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3 years ago