Question

A 2.450×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.3 ml . the density of water at 20.0∘c is 0.9982 g/ml. part a calculate the molality of the salt solution. express your answer to four significant figures and include the appropriate units.

Answer 1

Answer:The molality of the given solution is 0.0245 mol/kg.

Step-by-step explanation:

Let the dissolved or given mass of the NaCl = x grams

Final volume of water(solvent) after creating the sample = 1.000 L

`Molarity=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}* \text{Volume of solvent in liters}}`

`2.450* 10^(-2) mol/L=(x grams)/(58.44 g/mol* 1.000 L)`

x = 1.4317 g

The density of water at 20 C = 0.998 g/mL

`Density=\frac{\text{mass}}{\text{volume of water}}=0.998 g/mL=\frac{\text{mass of the water}}{1.000* 1000 mL}`

1 L = 1000 mL

Mass of the water(solvent) = 988 g = 0.988 kg(1 kg=1000 g)

`Molality=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}* \text{weight of the solvent in kg}}`

`Molality=(1.4317 g)/(58.44 g/mol* 0.998 kg)=0.0245 mol/kg`

The molality of the given solution is 0.0245 mol/kg.

Answer 2

Answer : The molality of the given solution is 0.02453 mole/kg.

Solution :

First we have to calculate the mass of NaCl (solute).

`Molarity=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}* \text{Volume of solution in liters}}`

`2.450* 10^(-2)mole/L=\frac{\text{Molar mass of NaCl}}{58.44g/mole* 1.000 L}`

`\text{Molar mass of NaCl}=1.43g`

Now we have to calculate the mass of water (solvent).

`Density=\frac{\text{Mass of water}}{\text{Volume of water}}`

`0.9982g/mL=\frac{\text{Mass of the water}}{999.3mL}`

Mass of the water (solvent) = 997.5 g

Now we have to calculate the molality of solution.

`Molality=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}* \text{weight of the solvent in kg}}`

`Molality=(1.43g* 1000)/(58.44g/mol* 997.5Kg)=0.02453mol/kg`

Therefore, the molality of the given solution is 0.02453 mole/kg.