Question

how many ml of a 0.216M aqueous solution of barium sulfide must be taken to obtain 12.5 grams of the salt?

Answer 1

Hey there!:

Molar mass BaS => 169.39 g/mol

Number of moles:

n = mass of solute / volume ( L )

n = 12.5 / 169.39

n = 0.0737 moles of BaS

Volume of solution:

M = n / V

0.216 = 0.0737 / V

V = 0.0737 / 0.216

V = 0.3412 L

Now converts the volume to mL :

0.3412 L * 1000 => 341.2 mL

Hope that helps!