How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH₄NO₃ in order to prepare a 0.452 m solution A) 769 g B) 36.2 g C) 100. g D) 0.157 g E) 157 g

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Quizplus · Accepted Answer

Molality (m) is defined as the number of moles of solute per kilogram of solvent. The molecular formula for ammonium nitrate is NH4NO3. The molar mass of NH4NO3 is calculated as follows: N (14.01) * 2 + H (1.01) * 4 + O (16.00) * 3 = 80.05 g/mol. The number of moles of NH4NO3 is 27.8 g / 80.05 g/mol = 0.347 moles. Given a molality of 0.452 m, which means 0.452 moles of NH4NO3 per kg of water, we can find the mass of water needed by rearranging the formula for molality: mass of water (kg) = moles of solute / molality. Thus, mass of water = 0.347 moles / 0.452 mol/kg = 0.768 kg, which is 768 g.

How Many Grams of Water Are Needed to Dissolve 27