Question

An aqueous solution contains 3.7 % NaCl by mass.

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 :

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

where,

n = moles of solute $NaCl=\frac{\text {given mass}}{\text {Molar mass}}=\frac{3.7}{58.5}=0.06$

$W_s$ = weight of solvent in g

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

$Molality=\frac{0.6moles\times 1000}{96.3g}=0.62mole/kg$

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)=${\text {given mass}}{\text {molar mass}}=\frac{96.3}{18}=5.35$

Total moles =0.6+5.35=5.95

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

Answer 2

Answer:

- Molality:

$m=0.65m$

- Mole fraction:

$x_{NaCl}=0.0116\\x_{H_2O}=0.9884$

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:

$m_{NaCl}=\frac{100g*3.7\%}{100\%} =3.7g$

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

$n_{NaCl}=\frac{3.7g}{58.45g/mol} =0.063molNaCl\\m_{H_2O}=100g-3.7g=96.3g*\frac{1kg}{1000g}=0.0963kgH_2O$

With which the molality results:

$m=\frac{n_{NaCl}}{m_{H_2O}}=\frac{0.063molNaCl}{0.0963kg}=0.65m$

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

$n_{H_2O}=\frac{96.3g}{18g/mol} =5.35molH_2O$

Thus, the mole fraction of sodium chloride is:

$x_{NaCl}=\frac{0.063}{5.35+0.063} =0.0116\\x_{H_2O}=\frac{5.35}{5.35+0.063} =0.9884$

Best regards.