# 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...?

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 $1 \cdot m L$ volume, which has a mass of $1 \cdot m L \times 1.33 \cdot g \cdot m {L}^{-} 1 = 1.33 \cdot g$.

For a $1 \cdot m L$ volume, there are..........

$10.5 \cdot m o l \cdot {L}^{-} 1 \times 1 \times {10}^{-} 3 \cdot L \times 40.0 \cdot g \cdot m o {l}^{-} 1 = 0.420 \cdot g$

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?