Respuesta :
Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate [tex]NH_4NO_3[/tex] 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 :
[tex]Molality=\frac{n\times 1000}{W_s}[/tex]
where,
n= moles of solute
Moles of [tex]NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles[/tex]
[tex]W_s[/tex] = weight of the solvent in g = ?
[tex]0.452=\frac{0.348\times 1000}{W_s}[/tex]
[tex]W_s=770g[/tex]
Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate [tex]NH_4NO_3[/tex] 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 =?
[tex]mole \: = \frac{mass}{molar \: mass} \\ \\ = \frac{27.8}{80} [/tex]
Mole of NH₄NO₃ = 0.3475 mole
- Finally, we shall determine the mass of water needed to prepare the solution.
Mole of NH₄NO₃ = 0.3475 mole
Molality of NH₄NO₃ = 0.452 M
Mass of water = ?
[tex]mass \: of \: water \: = \frac{mole}{molality} \\ \\ mass \: of \: water \: = \frac{0.3475}{0.452} [/tex]
Mass of water = 0.7688 Kg
Multiply by 1000 to express in grams
Mass of water = 0.7688 × 1000
Mass of water = 768.8 g
Thus, the mass of water needed to prepare the solution is 768.8 g
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