the mole fraction of a solute in aqueous solution of NaOH having molality one is
1)0.042
2)0.095
3)0.018
4)0.032
Answers
Answer:
The mole fraction of solvent is 0.95
Explanation:
We are given:
Molality of NaOH solution = 3m
This means that 3 moles of NaOH is present in 1 kg (1000 g) of solvent.
As, the solution is aqueous in nature which means that the solvent is water.
Calculating the moles of water in the solution, by using the equation:
\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}Number of moles=
Molar mass
Given mass
Given mass of water = 1000 g
Molar mass of water = 18 g/mol
Putting values in above equation , we get:
\text{Moles of water}=\frac{1000g}{18g/mol}=55.55molMoles of water=
18g/mol
1000g
=55.55mol
To calculate the mole fraction of solvent, we use the equation:
\chi_{water}=\frac{n_{water}}{n_{water}+n_{NaOH}}χ
water
=
n
water
+n
NaOH
n
water
We are given:
\begin{lgathered}n_{water}=55.55mol\\n_{NaOH}=3mol\end{lgathered}
n
water
=55.55mol
n
NaOH
=3mol
Putting values in above equation, we get:
\chi_{water}=\frac{55.55}{55.55+3}=0.95χ
water
=
55.55+3
55.55
=0.95
Hence, the mole fraction of solvent is 0.95