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What volume (in mL) of 0.415M silver nitrate will be required to precipitate as silver bromide, all bromide ions in 35.0 mL of 0.064M of calcium bromide?



2AgNO3(aq) + CaBr2(aq) → 2AgBr(s) + Ca(NO3)2(aq)

a
3.04 mL
b
10.8 mL
c
21.6 mL
d
43.2 mL


Sagot :

Answer:

im pretty sure its d

Explanation:

The volume (in mL) of 0.415M silver nitrate will be required to precipitate as silver bromide is 10.8ml , option B is the correct answer.

What is Molarity ?

Molarity is defined as the amount of solute (in moles)in per litre of solution.

It is also known as molar concentration of a solution , It is expressed in mol/l .

[tex]\rm M=\dfrac{n}{V}[/tex]

where n is the number of moles

and V is the volume in litres.

We can rearrange this equation to get the number of moles:

n= M * V

The molarity of calcium bromide is 0.064 M and the volume given is 35 ml,

the moles will be n = 0.064 * 35 = 2.24 moles

2AgNO₃(aq) + CaBr₂(aq) → 2AgBr(s) + Ca(NO₃)₂(aq)

From the reaction we can see that for every 1 mole of Calcium Bromide 2 moles of Silver Nitrate is needed , so for 2.24 moles 4.48 moles will be needed.

The molarity is given as 0.415 M and the number of moles is 4.48 so the volume required can be calculated from the above formula

[tex]\rm V=\rm\dfrac{n}{M} \\\\V=\rm\dfrac{4.48}{0.415} \\\\V= 10.8 ml\\[/tex]

Hence the volume (in mL) of 0.415M silver nitrate will be required to precipitate as silver bromide is 10.8ml , option B is the correct answer.

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