Answer:
[tex]m_{NaOH}=26.7gNaOH[/tex]
Explanation:
Hello,
In this case, given the molality of the solution which is 2.88 m (mol/kg) we can compute the moles of sodium hydroxide that are dissolved in the 232 grams of water (0.232 kg) as follows:
[tex]m=\frac{n}{m_{solvent}} \\\\n=m*m_{solvent}=2.88\frac{molNaOH}{kg}*0.232kg =0.668molNaOH[/tex]
Then, by using the molar mass of sodium hydroxide we compute the grams:
[tex]m_{NaOH}=0.668molNaOH*\frac{39.998gNaOH}{1molNaOH} \\\\m_{NaOH}=26.7gNaOH[/tex]
Regards.
