1.how many milliliters of an aqueous solution of 0.157 m aluminum bromide is needed to obtain 13.4 grams of the salt?
2.in the laboratory you dissolve 20.2 g of sodium nitrate in a volumetric flask and add water to a total volume of 375 . ml.

what is the molarity of the solution?

The answer to your question is:

1.- 320 ml

2.- 0.63 M

Explanation:

Data

V = ?   AlBr₃ = 0.157 M

mass = 13.4 g

Process

MW ALBr₃ = 27 + (80x3) = 267 g

                                267 g 1 mol

                                13.4 g    x

                                x = (13.4 x 1) / 267

                               x = 0.05 moles

Volume = 0.05/ 0.157

             = 0.320 l or 320 ml

2.-

NaNO₃ mass = 20.2 g

V = 375 ml

Molarity = ?

MW = 23 + 14 + 48 = 85g of NaNO₃

                          85 g of NaNO₃   1 mol

                          20.2 g                 x

                         x = (20.2 x 1) / 85

                         x = 0.24 mol

Molarity = 0.24 / 0.375

             = 0.63                

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