What is the mass percent of a solution that is #10.5*mol*L^-1# with respect to #NaOH#, and for which #rho_"solution"=1.33*g*mL^-1#...?

1 Answer
Jul 9, 2017

Answer:

We want #"Mass of caustic of soda"/"Mass of solution"xx100%#....

We get #"%NaOH(m/m)"-=32%#

Explanation:

Given that we have the molar concentration, and the solution density, we can fairly easily calculate the given quotient......

We work out the quotient on the basis of a #1*mL# volume, which has a mass of #1*mLxx1.33*g*mL^-1=1.33*g#.

For a #1*mL# volume, there are..........

#10.5*mol*L^-1xx1xx10^-3*Lxx40.0*g*mol^-1=0.420*g#

And thus #%"mass"=(0.420*g)/(1*mLxx1.33*g*mL^-1)xx100%=#

#=31.6%m/m#.

What safety precautions MUST BE observed when you make such a solution?