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Chemistry 23 Online
OpenStudy (anonymous):

(1)how many grams of NaCl are needed to prepare 1000g of 0.9% saline solution? (2)what is the% by mass of a solution that contains 50 grams of salt dissolved in 200 grams of solution? what is the concentration in ppm?

OpenStudy (aaronq):

1000 g x 0.9 % = ?

OpenStudy (aaronq):

what is the% by mass of a solution that contains 50 grams of salt dissolved in 200 grams of solution? 200 g + 50 g = 250 g 50g/250 g x100 %=

OpenStudy (anonymous):

900

OpenStudy (anonymous):

500

OpenStudy (aaronq):

lol nope

OpenStudy (anonymous):

why not?

OpenStudy (aaronq):

0.9 % = 0.009

OpenStudy (anonymous):

can you explain me please?

OpenStudy (aaronq):

1000 g x 0.9 % = 1000g x 0.9/100 =

OpenStudy (anonymous):

1000*0.9=900

thomaster (thomaster):

it's 0.9% percent means per hunderd so you have to divide it by 100 900/100 = 9 you would need 9 gram NaCl to make a 1000g 0.9% salt solution

OpenStudy (anonymous):

1000*0.9/100=9

thomaster (thomaster):

yes :)

OpenStudy (anonymous):

I think I can't pass chemistry class. I have a hard time to doing this type of math

thomaster (thomaster):

It's actually very easy 1000g = 100% so 900g = 90% 90g = 9% 9g = 0.9%

thomaster (thomaster):

If you think like this it's a lot easier

OpenStudy (anonymous):

Thanks for helping me.

thomaster (thomaster):

You need help with the second question?

OpenStudy (anonymous):

I got some idea now

OpenStudy (anonymous):

i will try to do by myself if i can't understand i will back

thomaster (thomaster):

hint: 50 grams dissolved in 200g is the same as 25g in 100g its easier to calculate with 100g as you can take that 100%

OpenStudy (anonymous):

thank you i will concentrate

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