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snow_tiger [21]
3 years ago
15

A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH4NO3 and 4.42 g of (NH4)3PO4 in enough

water to make 22.0 L solution. What is the molarity of NH4+ in the solution?
Chemistry
1 answer:
motikmotik3 years ago
8 0

Answer:

0.00725 M

Explanation:

Considering for NH_4NO_3

Mass = 5.66 g

Molar mass of NH_4NO_3 = 80.043 g/mol

Moles = Mass taken / Molar mass

So,  

<u>Moles = 5.66 / 80.043 moles = 0.0707 moles</u>

NH_4NO_3 will dissociate as:

NH_4NO_3\rightarrow NH_4^++NO_3^-

Thus 1 mole of NH_4NO_3 yields 1 mole of ammonium ions. So,

<u>Ammonium ions furnished by NH_4NO_3 = 1 × 0.0707 moles = 0.0707 moles</u>

Considering for (NH_4)_3PO_4

Mass = 4.42 g

Molar mass of (NH_4)_3PO_4 = 149.09 g/mol

Moles = Mass taken / Molar mass

So,  

Moles = 4.42 / 149.09 moles = 0.0296 moles

(NH_4)_3PO_4 will dissociate as:

(NH_4)_3PO_4\rightarrow 3NH_4^++PO_4^{3-}

Thus 1 mole of NH_4NO_3 yields 3 moles of ammonium ions. So,

<u>Ammonium ions furnished by (NH_4)_3PO_4 = 3 × 0.0296 moles = 0.0888 moles</u>

<u>Total moles of the ammonium ions = 0.0707 + 0.0888 moles = 0.1595 moles</u>

Given that:

Volume = 22.0 L

So, Molarity of the NH_4^+ is:

<u>Molarity = Moles / Volume = 0.1595 / 22 M = 0.00725 M</u>

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