How much water in kilogram is required to prepare 2.5 molar solution containing 20.0g of NaHCO3?

Ca= 40.08
Cl= 35.45
Na=22.99
H=1.01
C=12.01
O=16.00
S=32.07

1 Answer
Jan 26, 2018

The terms of this question are unclear....

Explanation:

We require that #((20.0*g)/(84.01*g*mol^-1))/("Volume of solution")=2.50*mol*L^-1#..

And so #"volume of solution"=((20.0*g)/(84.01*g*mol^-1))/(2.50*mol*L^-1)#

#0.0952*L#, i.e. a volume of #95.2*mL# of water is required....and thus a MASS of water #-=0.0952*kg#....