Question

4 g of naoh is contained in one decilitre of a solution. calculate mole fraction, molality , molarity of naoh (density of naoh =1.038)

Answer 1

hope it helps u :)
u can see pictures

Answer 2

Answer : The mole fraction, molality and molarity of NaOH are, 0.0177, 1.002 mole/Kg and 1 mole/L respectively.

Solution : Given,

Mass of NaOH (solute) = 4 g

Volume of solution = 1 deciliter = 100 ml

Density = 1.038 g/ml

• Calculation for mole fraction of NaOH

First we have to calculate the mass of solution.

$\text{Mass of solution}=Density\times Volume=1.038g/ml\times 100ml=103.8g$

Now we have to calculate the mass of solvent.

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}=103.8-4=99.8g$

Now we have to calculate the moles of solute, NaOH and solvent, $H_2O$

$\text{Moles of NaOH}=\frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}}=\frac{4g}{40g/mole}=0.1moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{99.8g}{18g/mole}=5.54moles$

Now we have to calculate the mole fraction of NaOH.

$\text{Mole fraction of NaOH}=\frac{\text{Moles of NaOH}}{\text{Moles of NaOH}+\text{Moles of }H_2O}=\frac{0.1}{0.1+5.54}=0.0177$

• Calculation for molality

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}$

$Molality=\frac{4g}\times 1000}{40g/mole\times 99.8Kg}=1.002mole/Kg$

• Calculation for molarity

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in L}}$

$Molarity=\frac{4g\times 1000}{40g/mole\times 100L}=1mole/L$

Therefore, the mole fraction, molality and molarity of NaOH are, 0.0177, 1.002 mole/Kg and 1 mole/L respectively.