Fundamentals of chem: Dilution G9

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Vanessa Romero-Campos 2B
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Fundamentals of chem: Dilution G9

Postby Vanessa Romero-Campos 2B » Thu Oct 01, 2015 8:16 pm

The question states: A chemist studying the properties of photographic emulsions needed to prepare 500.0 mL of 0.179 M AgN03 (aq). What mass of Silver Nitrate must be placed into a 500.0-mL volumetric flask, dissolved, and diluted to the mark with water?

I know I have to find the mass of AgNO3, but once I have the mass how do I use all these numbers to figure out the specific amount of silver nitrate that's needed? Do I divide the mass by 0.179 M? How do I use the 500 mL? Is there a specific formula I need?

P.S. This problem might be really simple, but this is my first time taking chemistry and I'm still not sure how to approach problems like these.

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Re: Fundamentals of chem: Dilution G9

Postby Chem_Mod » Thu Oct 01, 2015 8:32 pm

The first thing to know is that the molarity M = moles/liter. Use conversions that will get you towards the unit you want in the end, which is mass.

start with 0.179 moles/liter
multiply by 0.500 liters to cancel liters and get the moles AgNO3
multiply by molar mass of AgNO3 (g/mol) to get grams AgNO3

Aman Sankineni 2L
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Re: Fundamentals of chem: Dilution G9

Postby Aman Sankineni 2L » Wed Oct 02, 2019 3:49 pm

.179 M x .5 L = 0.0895 mol
AgN03 molar mass = (107.87 + 14 + 16*3) = 169.87 g mol^-1
.0895 mol x 169.87 g mol^-1 = 15.034 grams

Kassidy Ford 1I
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Re: Fundamentals of chem: Dilution G9

Postby Kassidy Ford 1I » Thu Oct 03, 2019 11:10 am

This problem confuses me too, why wouldn't you use the equation M1V1=M2V2? Is it because you are trying to find the mass?

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Re: Fundamentals of chem: Dilution G9

Postby 805307623 » Thu Oct 03, 2019 1:54 pm

I was also confused as to why the dilution formula could not be applied to this example!

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