11 months ago

Please assist me with chemistry assessment

Chemistry

1 answer:

Lera25 [3.4K]11 months ago

7 0

Answer:

I don't know

Explanation:

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Answer:

neutralization

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2 years ago

A 51.24-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be d

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Answer:

0.40 L

Explanation:

Calculation of the moles of $Ba(OH)_2$ as:-

Mass = 51.24 g

Molar mass of $Ba(OH)_2$ = 171.34 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{51.24\ g}{171.34\ g/mol}$

$Moles= 0.2991\ mol$

Volume = 1.20 L

The expression for the molarity is:

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

$Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M$

Thus,

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$Molarity_{working\ solution}=0.100\ M$

$Volume_{working\ solution}=1\ L$

$Volume_{stock\ solution}=?$

$Molarity_{stock\ solution}=0.24925\ M$

So,

$0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}$

$Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L$

The volume of 0.24925M stock solution added = 0.40 L

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