Question

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?

Answer 1

The answer to your question is:

1.- 320 ml

2.- 0.63 M

Step-by-step 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