Question

a 0.25 molar aqueous solution of NAOH is found to have density of 1.26 g/cc find molality and mole fraction

Answer 1

Answer:

Molality=0.2 And mole fraction =0.00358

Explanation:

Using formula

m=(1000M)/1000d-MMb

Where m is molality , M is molarity , d is density , Mb is molar mass of NAOH

Answer 2

Answer:

The molality of solution is equal to 0.2m and mole fraction of NaOH is equal to 0.00358.

Explanation:

Given: the concentration of NaOH  $=0.25M$

Density of aqueous solution, $d=1.26gml^{-1}$

$Density=\frac{Mass}{Volume}$

$1.26=\frac{m}{1000}$

Mass of the solution,$m_{solution}=1260g$

From 0.25M concentration, we can say that 0.25  moles of NaOH in 1L solution.

Therefore, number of moles of $NaOH=0.25moles$

Molar mass of NaOH =40gmol⁻¹

Mass of NaOH (solute)$=0.25*40=10g$

Mass of solvent $=1260-10=1250g$

Molality = (moles of NaOH)/[mass of solvent (kg)]

Molality, $m=\frac{0.25}{1250*10^{-3}}$

$Molality=0.2m$

Now, Number of moles of water (solvent)  $=\frac{1250}{18}=69.44mol$

Mole fraction of NaOH $=\frac{0.25}{0.25+69.44}$

                                      $=\frac{0.25}{69.69}$

Mole fraction of NaOH $=0.00358$