Question

Calculate the mole fraction, molarity and molality of NH3 if it is in
solution composed of 30.6 g NH3 in 81.3 g of H2O. The density of the
solution is 0.982 g/mL and the density of water is 1.00 g/mL​

Answer 1

Answer:

First of all let's calculate the molar weight of

NH3 =14+ 3= 17g

H2O =2 + 16= 18g

Now let's calculate the number of moles of

NH3 in 30.6g by dividing it by molar weight = 30.6/17 = 1.8 moles

Similarly, the number of moles of H20 in 81.3g = 81.3/18 = 4.5

Mole fraction = number of moles of solute/number of moles of solvent = 1.8/4.5 = 0.4

Molality = number of moles of solute /weight of solvent in kg = 1.8/81.3 X 1000 = 22 moles per gm of water

Molarity = number of moles of solute / volume of solvent in L = 1.8/81.3 × 1000 = 22 M per mL of water

Now except for the units, the molarity and molality is the same in this case, this is because, density of water is 1g per mL at room temperature

Step-by-step explanation:

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Answer 2

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First of all let's calculate the molar weight of

NH3 =14+ 3= 17g

H2O =2 + 16= 18g

Now let's calculate the number of moles of

NH3 in 30.6g by dividing it by molar weight = 30.6/17 = 1.8 moles

Similarly, the number of moles of H20 in 81.3g = 81.3/18 = 4.5

Mole fraction = number of moles of solute/number of moles of solvent = 1.8/4.5 = 0.4

Molality = number of moles of solute /weight of solvent in kg = 1.8/81.3 X 1000 = 22 moles per gm of water

Molarity = number of moles of solute / volume of solvent in L = 1.8/81.3 × 1000 = 22 M per mL of water

Now except for the units, the molarity and molality is the same in this case, this is because, density of water is 1g per mL at room temperature.