Question

mole fraction of solvent in aqueous solution of NaOH having molality of 3 is

Answer 1

Answer:0.95

Explanation:

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Answer 2

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:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation , we get:  

$\text{Moles of water}=\frac{1000g}{18g/mol}=55.55mol$

To calculate the mole fraction of solvent, we use the equation:

$\chi_{water}=\frac{n_{water}}{n_{water}+n_{NaOH}}$

We are given:

$n_{water}=55.55mol\\n_{NaOH}=3mol$

Putting values in above equation, we get:

$\chi_{water}=\frac{55.55}{55.55+3}=0.95$

Hence, the mole fraction of solvent is 0.95