Asked by Fai
Find the number of mg Na2co3 FW = 106g/mol required to prepare 500ml of 9.2ppm NA solution.
ppm = mg/L
9.2mg/1000ml
x= 0.0092 x 500ml = 4.6mg x 23 = 106mg
The right answer 10.6mg
NA weight = 23
Who helps me to solve it for me.
ppm = mg/L
9.2mg/1000ml
x= 0.0092 x 500ml = 4.6mg x 23 = 106mg
The right answer 10.6mg
NA weight = 23
Who helps me to solve it for me.
Answers
Answered by
DrBob222
1 ppm = 1 mg/L
9.2 ppm = 9.2 ppm x 0.500L = 4.6 mg Na.
Convert that to mg Na2CO3 by
4.6 mg x (106/23) = ? mg Na2CO3. Dissolve in water and make up to 0.500 L.
9.2 ppm = 9.2 ppm x 0.500L = 4.6 mg Na.
Convert that to mg Na2CO3 by
4.6 mg x (106/23) = ? mg Na2CO3. Dissolve in water and make up to 0.500 L.
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