Question

A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH₄NO₃ and 4.42 g of (NH₄)₃ PO₄ in enough water to make 20.0 L of solution. What are the molarities of NH₄⁺ and of PO₄³⁻ in the solution?

Answer 1

In finding the molarity of a solution, we use the following formula:

$M=moles soluteL solution$

What is Molarity?

The number of moles of the solute is calculated by dividing the mass of the solute by its molar mass.

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The molar mass of  NH4NO3 and (NH4)3PO4 are  80.043 g/mol and 149.0867 g/mol, respectively.

$molesNH+4inNH4NO3=5.66 g80.043 g/mol×1molNH+41molNH4NO3=0.0707 mol$

$molesNH+4in(NH4)3PO4=4.42 g149.0867 g/mol×3molNH+41mol(NH4)3PO4=0.0889 mol$

$total molesNH+4=0.0707 mol+0.0889 mol=0.1596 mol$

$molesPO3−4in(NH4)3PO4=4.42 g149.0867 g/mol×1molPO3−41mol(NH4)3PO4=0.0296 mol$

$[NH+4]=0.1596 mol20.0 L=7.98×10−3 M NH+4$

$[PO3−4]=0.0296 mol20.0 L=1.48×10−3 M PO3−4$

Therefore, $[PO3−4]$ has a molarity of  $1.48×10−3 M PO3−$

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