Question

how many grams of sodium bromide are there in 55.0mL of a 1.90M aqueous solution? please answer as soon as possible and show work thank you

Answer 1

We know that Molarity , M = number of moles / volume of solution in L

So number of moles of NaBr , n = Molarity x Volume in L

                                               = 1.90 M x (55.0/1000) L

                                              = 0.1045 moles

Molar mass of NaBr = At.mass of Na + At.mass of Br

                              = 23+80

                              = 103 g/mol

We know that number of moles = mass/molar mass

So mass of NaBr , m = number of moles x molar mass

                                = 0.1045 mol x 103 (g/mol)

                                = 10.8 g

Therefore the mass of sodium bromide is 10.8 g