Question

An aqueous solution contains 3.7 % NaCl by mass.

Calculate the molality of the solution.
Calculate the mole fraction of the solution.

Answer 1

Answer : The molality of solution will be 0.62m and the mole fraction of solute is 0.10

Solution:

Molar mass of
`NaCl` = 58.5g/mole

3.7%
`NaCl` solution means 3.7 g of
`NaCl` is present in 100 g of solution

Mass of water (solvent) =(100-3.7)=96.3 g

Molality : It is defined as the number of moles of solute present per kg of solvent.

Formula used :

`Molality=(n* 1000)/(W_s)`

where,

n = moles of solute
`NaCl=\frac{\text {given mass}}{\text {Molar mass}}=(3.7)/(58.5)=0.06`

`W_s` = weight of solvent in g

Now put all the given values in the formula of molality, we get

`Molality=(0.6moles* 1000)/(96.3g)=0.62mole/kg`

Therefore, the molality of solution will be 0.62 mole/kg.

Mole fraction of solute(NaCl)=
`\text {number of moles of solute}}{\text {total moles}}`

Moles of solute(NaCl)= 0.6

Moles of solvent (water)=
`{\text {given mass}}{\text {molar mass}}=(96.3)/(18)=5.35`

Total moles =0.6+5.35=5.95

Mole fraction of solute(NaCl)=
`{0.6}{5.95}=0.10`

Answer 2

- Molality:

`m=0.65m`

- Mole fraction:

`x_(NaCl)=0.0116\\x_(H_2O)=0.9884`

Step-by-step explanation:

Hello,

- Molality: in this case, the solution is formed by 3.7% by mass of sodium chloride and water, thereby, we assume we have 100 g of solution in order to find the mass of sodium chloride as shown below:

`m_(NaCl)=(100g*3.7\%)/(100\%) =3.7g`

Hence, the moles of sodium chloride and kilograms of water are:

`n_(NaCl)=(3.7g)/(58.45g/mol) =0.063molNaCl\\m_(H_2O)=100g-3.7g=96.3g*(1kg)/(1000g)=0.0963kgH_2O`

With which the molality results:

`m=(n_(NaCl))/(m_(H_2O))=(0.063molNaCl)/(0.0963kg)=0.65m`

- Mole fraction: in this case, since we have previously computed the moles of sodium chloride, we now compute the moles of water as:

`n_(H_2O)=(96.3g)/(18g/mol) =5.35molH_2O`

Thus, the mole fraction of sodium chloride is:

`x_(NaCl)=(0.063)/(5.35+0.063) =0.0116\\x_(H_2O)=(5.35)/(5.35+0.063) =0.9884`

Best regards.