Question

How many grams of solute are present in 735 mL of 0.380 M KBr?

Answer 1

volume , V = 735 mL

= 0.735 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.38*0.735

= 0.2793 mol

Molar mass of KBr = 1*MM(K) + 1*MM(Br)

= 1*39.1 + 1*79.9

= 119 g/mol

we have below equation to be used:

mass of KBr,

m = number of mol * molar mass

= 0.2793 mol * 119 g/mol

= 33.2 g

Answer:  33.2 g