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

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?