You need to make an aqueous solution of 0.177 M chromium(III) nitrate for an experiment in lab, using a 250 mL volumetric flask. How much solid chromium(III) nitrate should you add

10.53 g of Cr(NO₃)₃

Explanation:

From the question given above, the following data were obtained:

Molarity of Cr(NO₃)₃ = 0.177 M

Volume of solution = 250 mL

Mass of Cr(NO₃)₃ =?

Next, we shall convert 250 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

250 mL = 250 mL × 1 L / 1000 mL

250 mL = 0.25 L

Thus, 250 mL is equivalent to 0.25 L.

Next, we shall determine number of mole of Cr(NO₃)₃ in the solution. This can be obtained as follow:

Molarity of Cr(NO₃)₃ = 0.177 M

Volume of solution = 0.25 L

Mole of Cr(NO₃)₃ =?

Molarity = mole / Volume

0.177 = Mole of Cr(NO₃)₃ / 0.25

Cross multiply

Mole of Cr(NO₃)₃ = 0.177 × 0.25

Mole of Cr(NO₃)₃ = 0.04425 mole

Finally, we shall determine the mass of Cr(NO₃)₃ needed to produce the solution. This can be obtained as follow:

Mole of Cr(NO₃)₃ = 0.04425 mole

Molar mass of Cr(NO₃)₃ = 52 + 3[14 + (3×16)]

= 52 + 3[14 + 48]

= 52 + 3[62]

= 52 + 186

Molar mass of Cr(NO₃)₃ = 238 g/mol

Mass of Cr(NO₃)₃ =?

Mole = mass / Molar mass

0.04425 = Mass of Cr(NO₃)₃ / 238

Cross multiply

Mass of Cr(NO₃)₃ = 0.04425 × 238

Mass of Cr(NO₃)₃ = 10.53 g

Thus, 10.53 g of Cr(NO₃)₃ should be added to the 250 mL and fill with water to the mark.

c. electrical charges of ionic bonds.