Question
Asked Oct 14, 2019
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Question 5

4. A student added 25.00 grams of sodium nitrate to a 2.00 Litre volumetric flask. What
is the molarity (M) of the resulting solution?
5. A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and
50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity
(M) of the resulting solution relative to the nitrate ion, NO31-?
6. How many grams of phosphoric acid are required to react with 50.00 mL of a 0.800
M calcium nitrate solution?
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4. A student added 25.00 grams of sodium nitrate to a 2.00 Litre volumetric flask. What is the molarity (M) of the resulting solution? 5. A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO31-? 6. How many grams of phosphoric acid are required to react with 50.00 mL of a 0.800 M calcium nitrate solution?

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Expert Answer

Step 1

The molarity of nitrate ion can be calculated by first calculating the moles of nitrate present in the solution.

Calculating the number of moles of the calcium nitrate in 20 grams.

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Mass No. of moles Molar mass 20 No. of moles of Ca (NO3)2 = 0.121 moles 164 g/mol 1 mole of Calcium nitrate consists of two moles of nitrate ions, Therefore, 0.121 moles will contain = 0.121 x 2 = 0.242 moles of nitrate ion.

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Step 2

Now the number of moles of sodium nitrate can be calculated as,

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Mass No. of moles = Molar mass 10 No. of moles of N2NO3 0.117 moles 85 g/mol 1 mole of NaNO3 consists of 1 mole of nitrate ion, Therefore 0.117 moles of sodium nitrate will contain = 0.117 moles of nitrate ion.

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Step 3

The number of moles of aluminium nitrate (Al(NO3)...

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Mass No. of moles = Molar mass 50 No. of moles of Al(NO3); 0.234 moles 213 g/mol 1 mole of Al(NO3)3 consists of 3 moles of nitrate ion, Therefore, 0.234 moles of aluminium nitrate will contain = 0.234 x 3 moles of nitrate ion 0.702

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