Question #2a61c

1 Answer
Antoine · Media Owl
Apr 7, 2015

You can always use

#"moles" = "mass(g)"/("molar mass"("g "mol^-1))#

or #"Moles"# = Molarity#("mol "dm^-3)xx"Volume"(dm^3)#

Example:
Find the mass of 2 mol of NaOH?

Solution:
moles = #"mass"/"Molar mass"#

#=> mass = "moles"xx"molar mass"#

moles = #2# mol , molar mass = 23 + 16 + 1 = 40 g # mol^-1#

hence #" mass "= "2 mol "xx"40 g "mol^-1# = 80 g

This means that 2mol of #"NaOH"# is equivalent to a mass of #80" g"#

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