G7

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Kyle_Diep_4E
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

G7

Postby Kyle_Diep_4E » Fri Oct 02, 2015 8:10 pm

Hey! To be honest, I don't know where to even begin with this problem. If anyone could help me that would be nice!

You need to prepare 510. g of an aqueous solution containing 5.45% KNO3 by mass. Describe how you would prepare the solution and what mass of each component you would use.

Mieka McFarlane 2L
Posts: 10
Joined: Fri Sep 25, 2015 3:00 am

Re: G7

Postby Mieka McFarlane 2L » Fri Oct 02, 2015 9:30 pm

Ok so to start off we have to find out how much grams of KNO3 are in the sample of 510 g
The first step is to to get 5.45 from a percentage to a decimal so we divide it by 100 and it becomes .0545.
Then we multiply .0545 by the 510 g to get how much grams of KNO3 were in this aqueous solution which comes out to be 27.8g
Because it was an aqueous solution we know that the other component of the solution is water and because we know the total amount of the aqueous solution as well as the amount of grams of KNO3 we can find how much water is left (the remainder)
510g = 27.8g+x
x= 482.2 H20

Hadji Yono-Cruz 2L
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

Re: G7

Postby Hadji Yono-Cruz 2L » Tue Oct 16, 2018 4:02 pm

Mieka McFarlane 2L wrote:Ok so to start off we have to find out how much grams of KNO3 are in the sample of 510 g
The first step is to to get 5.45 from a percentage to a decimal so we divide it by 100 and it becomes .0545.
Then we multiply .0545 by the 510 g to get how much grams of KNO3 were in this aqueous solution which comes out to be 27.8g
Because it was an aqueous solution we know that the other component of the solution is water and because we know the total amount of the aqueous solution as well as the amount of grams of KNO3 we can find how much water is left (the remainder)
510g = 27.8g+x
x= 482.2 H20


This was really helpful thank you.


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