## Fundamentals of chem: Dilution G9

Vanessa Romero-Campos 2B
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Joined: Fri Sep 25, 2015 3:00 am

### Fundamentals of chem: Dilution G9

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.

Chem_Mod
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### Re: Fundamentals of chem: Dilution G9

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.

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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Joined: Fri Aug 30, 2019 12:17 am

### Re: Fundamentals of chem: Dilution G9

.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

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

805307623
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### Re: Fundamentals of chem: Dilution G9

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