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3 years ago

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?

1 answer:

5 0

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ 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 :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

Moles of $NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles$

$W_s$ = weight of the solvent in g = ?

$0.452=\frac{0.348\times 1000}{W_s}$

$W_s=770g$

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

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Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

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∴ P1 = 10 atm

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d) reversibly cooled at constant pressure:

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∴ P2 = P1

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