How many grams of solute do you need to prepare 353.2 mL of 0.407 M NaOH?

C=n/v
Concentration or molarity = number of moles / volume of the solution
……So we have the concentration and the volume….
that means we can find the number of moles using this formula C=n/v
rearrange this formula to make the number of moles the subject:
n=c times by v
c= 0.407 M
v= this has to be in litres so 0.3532 L

n= 0.3532 times by 0.407 = 0.1438 moles of NaOH
so to get the grams
n=m/M
m= n times by M
Na molar mass is 23, (from periodic table)
so, m= 23 times by 0.1438 = 3.306 grams of Na

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One Response to “How many grams of solute do you need to prepare 353.2 mL of 0.407 M NaOH?”

Unkownly FamoUs on October 1st, 2009 at 3:00 am

C=n/v
Concentration or molarity = number of moles / volume of the solution
……So we have the concentration and the volume….
that means we can find the number of moles using this formula C=n/v
rearrange this formula to make the number of moles the subject:
n=c times by v
c= 0.407 M
v= this has to be in litres so 0.3532 L

n= 0.3532 times by 0.407 = 0.1438 moles of NaOH
so to get the grams
n=m/M
m= n times by M
Na molar mass is 23, (from periodic table)
so, m= 23 times by 0.1438 = 3.306 grams of Na
References :

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