How many moles of potassium chloride are present in a #3L# volume of #1mol*L^-1# #KCl(aq)#?

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Answer 1 · 2017-06-03T14:06:48

Well, I gets a #3*"mole"# quantity of #KCl#.......

#"Molarity"="Moles of solute"/"Volume of solution"#.........

And thus for #"moles of solute"#, we take the product,

#"Molarity"xx"Volume of solution"#

#=3*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)#

#=3.0*mol#

Given that #"Avogadro's number"-=6.022xx10^23*mol^-1#, how many ATOMS are in this quantity?

How many moles of potassium chloride are present in a #3*L# volume of #1*mol*L^-1# #KCl(aq)#?