Question

How many grams of solute are present in 845 mL of 0.530 M KBr?

Answer 1

fitst find out the moles using the fromula

Molarity of KBr = no of moles of KBr / volume of the solvent in liters

Molarity of KBr = 0.530

volume = 845 mL convert in to liters = 845/1000 = 0.845 L

0.530 M = no of moles / 0.845 L

no of moles = 0.530 m/L x 0.845 L =0.44785 moles

now use the another fomula

no of moles = weight / molar mass

weight of KBr = no of moles of KBr x molar mass of KBr

= 0.44785 moles x 119.0 g/mol ( molar mass of KBr = 119.0 g/mol from google)

= 53.3 grams