Question

A student wishes to prepare 300-mL of a 0.180 M potassium bromide solution using solid potassium bromide, a 300-mL volumetric flask, and deionized water.
(a) How many grams of potassium bromide must the student weigh out?

______g


(b) Which of the following would NOT be an expected step in the procedure used by the student?

a. Dry the flask in a drying oven.

b. Carefully add water until the bottom of the meniscus sits on the mark on the neck of the flask.

c. Carefully transfer the salt sample to the volumetric flask.

Answer 1

Answer:

(a) 64.3 g

(b) c. Carefully transfer the salt sample to the volumetric flask.

Explanation:

(a)

A student wishes to prepare 300-mL (0.300 L) of a 0.180 M potassium bromide solution. The moles of KBr required are:

0.300 L Solution × (1.80 mol KBr/1 L Solution) = 0.540 mol

The molar mass of KBr is 119.0 g/mol. The mass corresponding to 0.540 moles is:

0.540 mol × (119.0 g/mol) = 64.3 g

(b)

c. Carefully transfer the salt sample to the volumetric flask.

The salt must be dissolved in a beaker prior to the transfer of this solution to the volumetric flask.

Answer 2

Answer:

a) 6.426  grams of potassium bromide student must weigh.

b) The correct answer is option a.

Explanation:

$Concentration(C)=\frac{Moles(n)}{Volume (L)}$

a) Concentration of potassium bromide = c = 0.180 M

Moles of potassium bromide = n

Volume of the  potassium bromide solution = 300 mL = 0.3 L

$0.180 M=\frac{n}{0.3 L}$

$n=0.180 M\times 0.3 L=0.054 mol$

Mass of 0.054 moles of potassium bromide :

= $0.054 mol\times 119 g/mol=6.426 g$

6.426  grams of potassium bromide student must weigh.

b) Drying of flask in drying oven. This is because solution will be prepared by adding water after adding solute. So drying the flask will be not required.