Question

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`...?

Answer 1

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?