Answer: 266.055 g
Explanation:
We have the following data related to this unknown substance:
[tex]grams-of-substance=29 g[/tex] are the grams of substance
[tex]molarity=0.285M[/tex] is the molarity of the substance, given by the following formula:
[tex]molarity=\frac{moles-of-solute}{Volume-of-solution-in-liters}[/tex] (1)
In this case the volume of solution is [tex]385 ml \frac{1L}{1000 mL}=0.385 L[/tex]
Finding the moles of solute:
[tex]moles-of-solute=(0.285M)(0.385 L)[/tex] (2)
[tex]moles-of-solute=0.109 moles[/tex] (3)
Now we have to convert this to grams by the following relation:
[tex]\frac{grams-of-substance}{moles-of-substance}=\frac{molar-mass-substance-in-grams}{1 mole}[/tex] (4)
Finding [tex]molar-mass-substance-in-grams[/tex]:
[tex]molar-mass-substance-in-grams=\frac{grams-of-substance}{moles-of-substance} 1 mole[/tex] (5)
[tex]molar-mass-substance-in-grams=\frac{29 g}{0.109} 1 mole[/tex] (6)
[tex]molar-mass-substance-in-grams=266.05 g[/tex]
