Question

Mole fraction of A in \[{{H}_{2}}O\] is 0.2 then the molality of A in \[{{H}_{2}}O\] is?

Answer 1

Hint: In the given numerical ‘A’ is the solute and water is the solvent. To find the molality of A in \[{{H}_{2}}O\], first, we will calculate the moles of water. Secondly, the number of moles of A will be calculated because it will be needed to find the molality of A.

Complete answer:
As we know that water is the solvent, so let’s assume the mass of water to be 1000 grams.
The molar mass of Hydrogen = 1
The molar mass of Oxygen = 16
$\therefore \text{ Molar mass of }{{\text{H}}_{2}}O\text{ = 2 }\times \text{ H + O}$
$=\text{ 2 }\times \text{ 1 + 16}$
$=\text{ 18}$
Thus,
$\Rightarrow No.\text{ of moles of solvent }{{\text{H}}_{2}}O\text{ = }\dfrac{Mass\text{ of }{{\text{H}}_{2}}O}{Molar\text{ mass of }{{\text{H}}_{2}}O}$
$=\text{ }\dfrac{1000}{18}$
$=\text{ 55}\text{.55 moles}$
Let the number of moles of A be x. Also, the given mole fraction is 0.2.
$\Rightarrow \text{ }Mole\text{ fraction = }\dfrac{\text{Number of moles of solute}}{No.\text{ of moles of solute + No}\text{. of moles of solvent}}$
Cross multiply 0.2 with the denominator
Now, Solve the bracket on the LHS
$0.2\text{ }x\text{ + 11}\text{.11 = x}$
Taking 0.2x to the RHS
$11.11\text{ = 1 x - 0}\text{.2 x}$
$11.11\text{ = 0}\text{.8 x}$
Taking 0.8 in the denominator
$x\text{ = }\dfrac{11.11}{0.8}$
Dividing 11.11 by 0.8, we get
$x\text{ = 13}\text{.88}$
Thus, the number of moles of solute A= 13.88.
Now,
The molality of the solution is given by ‘m’.
$\Rightarrow \text{ }Molality\text{ }\left( m \right)\text{ = }\dfrac{No.\text{ of moles of solute}}{Mass\text{ of solvent in kg}}$
We assumed the mass of the solvent as 1000 grams which is equal to 1kg.
$=\text{ }\dfrac{13.88}{1}$
$=\text{ 13}\text{.88}$
Therefore, the Molality of A in \[{{H}_{2}}O\]is 13.88 m.
Additional information:
Alternate method:
We know that the mole fraction of a solution is 1.
The mole fraction of solute A is 0.2.
$\therefore \text{ Mole fraction of }{{\text{H}}_{2}}O\text{ = 1- 0}\text{.2}$
$=\text{ 0}\text{.8}$
Moles of Solvent \[{{H}_{2}}O\] is 0.8.
Now, we will find the given mass of water.
$\Rightarrow \text{ }Weight\text{ of }{{\text{H}}_{2}}O\text{ = No}\text{. of moles of }{{\text{H}}_{2}}O\text{ }\times \text{ Molar mass of }{{\text{H}}_{2}}O$
$=\text{ 0}\text{.8 }\times \text{ 18}$
$=\text{ 14}\text{.4 grams}$
Thus, the weight of Solvent is 14.4 g.
But the weight of solvent has to be in kg to find the Molality. So the weight of the solvent has to be divided by 1000.
\[Weight\text{ }of\text{ }Solvent\text{ }in\text{ }kg\text{ }=\text{ }\dfrac{14.4}{1000}\]
$\Rightarrow Molality\text{ }\left( m \right)\text{ = }\dfrac{No.\text{ of moles of solute}}{\text{Weight of solvent in kg}}$
$=\text{ }\dfrac{0.2}{0.0144}$
$=\text{ 13}\text{.88 molal}$
Molality= 13.88 molal.

Note:
Always remember to check the Si units of the given quantities. Several times the weight is given in grams, so it has to be converted into kgs before using in the desired formula. You can also multiply the numerator by 1000 if you want to solve it one step and not find out the weight in kg separately.