What is the molality of a solution of phosphoric acid, #H_3PO_4# that contains 24.5 g of phosphoric acid (molar mass 98.0 g) in 100 g of #H_2O#?
The molality is 2.50 mol/kg.
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To find the molality of the solution:
 Calculate the moles of phosphoric acid using its mass and molar mass.
 Calculate the mass of water in kilograms.
 Use the formula for molality: ( \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ).
Given:
 Mass of phosphoric acid (( H_3PO_4 )): 24.5 g
 Molar mass of ( H_3PO_4 ): 98.0 g/mol
 Mass of water (( H_2O )): 100 g

Moles of ( H_3PO_4 ): [ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{24.5 , \text{g}}{98.0 , \text{g/mol}} = 0.25 , \text{mol} ]

Mass of water in kg: [ \text{mass of water (kg)} = \frac{100 , \text{g}}{1000} = 0.1 , \text{kg} ]

Molality: [ \text{molality} = \frac{0.25 , \text{mol}}{0.1 , \text{kg}} = 2.5 , \text{mol/kg} ]
The molality of the solution of phosphoric acid (( H_3PO_4 )) in water is 2.5 mol/kg.
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