Describe the preparation of 900 mL of 3.00 M HNO3 from the commercial reagent that is 70.5% HNO3 (w/w) and has a specific gravity of 1.42.
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1 point
Dilute 170 mL of the concentrated reagent to 0.900 mL with water.
Dilute 170 mL of the concentrated reagent to 900 mL with water.
Dilute 170 mL of the concentrated reagent to 1000 mL with water.
Dilute 170 mL of the concentrated reagent to 90 mL with water.
The correct answer is:
Dilute 170 mL of the concentrated reagent to 900 mL with water.
To prepare 900 mL of 3.00 M HNO3 solution, we need to calculate the amount of HNO3 in the 70.5% HNO3 reagent.
First, determine the mass of HNO3 in the 170 mL of the concentrated reagent:
170 mL x 0.705 = 119.85 g of HNO3
Next, calculate the moles of HNO3 in 119.85 g:
119.85 g / 63.01 g/mol = 1.903 moles
Now, use the formula M1V1 = M2V2 to dilute the solution:
(1.903 moles / 0.900 L) = (3.00 M / 0.900 L)
1.903 moles = 2.667 moles
So, you will need to dilute 170 mL of the concentrated reagent to a total volume of 900 mL with water to prepare 900 mL of 3.00 M HNO3.