Respuesta :
Answer:The molality of the given solution is 0.0245 mol/kg.
Explanation:
Let the dissolved or given mass of the NaCl = x grams
Final volume of water(solvent) after creating the sample = 1.000 L
[tex]Molarity=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}\times \text{Volume of solvent in liters}}[/tex]
[tex]2.450\times 10^{-2} mol/L=\frac{x grams}{58.44 g/mol\times 1.000 L}[/tex]
x = 1.4317 g
The density of water at 20 C = 0.998 g/mL
[tex]Density=\frac{\text{mass}}{\text{volume of water}}=0.998 g/mL=\frac{\text{mass of the water}}{1.000\times 1000 mL}[/tex]
1 L = 1000 mL
Mass of the water(solvent) = 988 g = 0.988 kg(1 kg=1000 g)
[tex]Molality=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}\times \text{weight of the solvent in kg}}[/tex]
[tex]Molality=\frac{1.4317 g}{58.44 g/mol\times 0.998 kg}=0.0245 mol/kg[/tex]
The molality of the given solution is 0.0245 mol/kg.
Answer : The molality of the given solution is 0.02453 mole/kg.
Solution :
First we have to calculate the mass of NaCl (solute).
[tex]Molarity=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}\times \text{Volume of solution in liters}}[/tex]
[tex]2.450\times 10^{-2}mole/L=\frac{\text{Molar mass of NaCl}}{58.44g/mole\times 1.000 L}[/tex]
[tex]\text{Molar mass of NaCl}=1.43g[/tex]
Now we have to calculate the mass of water (solvent).
[tex]Density=\frac{\text{Mass of water}}{\text{Volume of water}}[/tex]
[tex]0.9982g/mL=\frac{\text{Mass of the water}}{999.3mL}[/tex]
Mass of the water (solvent) = 997.5 g
Now we have to calculate the molality of solution.
[tex]Molality=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}\times \text{weight of the solvent in kg}}[/tex]
[tex]Molality=\frac{1.43g\times 1000}{58.44g/mol\times 997.5Kg}=0.02453mol/kg[/tex]
Therefore, the molality of the given solution is 0.02453 mole/kg.
