cascyrio2002
18.01.2021 •
Chemistry
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
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Ответ:
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.
Ответ:
144 grams
Step-by-step explanation:
Each crayon weighs 8 grams
8*18= 144