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?

The mass of water needed to dissolve 27.8 g of ammonium nitrate, NH₄NO₃ in order to prepare a 0.452 M solution is 768.8 g

We'll begin by calculating the number of mole in 27.8 g of ammonium nitrate, NH₄NO₃. This can be obtained as follow:

Mass of NH₄NO₃ = 27.8 g

Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16) = 80 g/mol

Mole of NH₄NO₃ = 0.3475 mole

Molality of NH₄NO₃ = 0.452 M

Mass of water = 0.7688 Kg

Multiply by 1000 to express in grams

Mass of water = 0.7688 × 1000

Thus, the mass of water needed to prepare the solution is 768.8 g

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770 g water are needed to dissolve 27.8 g of ammonium nitrate in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

where,

n= moles of solute

Moles of  

= weight of the solvent in g = ?

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate in order to prepare a 0.452 m solution

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).