Help with G9  [ENDORSED]

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Kyra LeRoy 1E
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

Help with G9

Postby Kyra LeRoy 1E » Mon Oct 02, 2017 10:23 pm

So for question G9 I'm using the formula M(I) x V(I) = M(F) x V(F)
I got
M(I)=.179mol/L
V(I)=.5L

but I can't figure out M2 and V2? or am I possibly mixing up which ones are the initial and which are final?
This might actually turn out to be simple, but I'm stuck! Help!

Thanks :)

Vivian Nguyen
Posts: 66
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

Re: Help with G9

Postby Vivian Nguyen » Mon Oct 02, 2017 10:31 pm

Hello Kyra,

For G9 they're asking you for the mass of silver nitrate needed to prepare that 500. mL of 0.179 M silver nitrate. You're going to want to use the other formula where Molarity = moles of solute (n) / volume of solution (v). In the question you're given the molarity (0.179 mol/L) and the volume ( .500 Liters), so you can solve for moles, and then solve for the mass. Hope this helps!

Ryan Sydney Beyer 2B
Posts: 82
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Re: Help with G9  [ENDORSED]

Postby Ryan Sydney Beyer 2B » Tue Oct 03, 2017 1:13 pm

For question G9 you don't actually use the M[i]V[i] = M[f]V[f]. Instead we'll be using the M = n/V.

In this problem they are asking for the mass of silver nitrate which needs to be added into the 500.0 ml flask for the molarity to equal 0.179 M. The way we'll find mass from the information given is manipulating the M = n/V equation so that we can solve for moles of silver nitrate and then convert moles to mass with the molar mass of silver nitrate.

So we plug in the information given into M = n/V and we end up with :: 0.179 M AgNO3 = (mol AgNO3) / (0.5000 L) [we had to convert 500.0 ml to liters because this equation operates easiest with the units of molarity, moles, and liters]

Solve for moles of AgNO3 :: (0.179 M AgNO3) * (0.5000 L) = mol AgNO3

= 0.0895 mol AgNO3

Then to convert moles of AgNO3 to grams we multiply by the molar mass of AgNO3 (169.87g/mol).

0.0895 mol AgNO3 * (169.87 g / 1 mol AgNO3) = 15.2 g AgNO3


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