Respuesta :

Willchemistry
Molar mass KBr = 119 g/mol

Volume in liters: 250.0mL / 1000 => 0.25 L

n = M x V

n = 1.20 x 0,25 => 0.3 moles of KBr

Therefore:

1 mole KBr ----------- 119 g
0.3 moles KBr -------- ??

0.3 x 119 / 1 => 35.7 g of KBr