Question

In aqueous solution of h2so4 the mole fraction of water is 0.85 .the molality of solution is?

Answer 1

Mole fraction of solution = solute + solvent = 1

Here, Mole fraction of water is 0.85 so mole fraction of sulphuric acid will be (1-0.85) = 0.15

Hence each mole of the solution, 0.15 mole of sulphuric acid is dissolved .

So mass of solvent (water) is ,

For 1 mole of water H2O = 18 gm

So, 0.85 mole of water = 18 x 0.85 = 15.3 gm =0.0153 Kg

Thus, Molality (m) = no. of moles of solute (Sulphuric acid)/ mass of solvent in Kg

Answer>> Hence, Molality of the solution (m) = 0.15 / 0.0153 = 9.8 m (nearly equal to 10)

Answer 2

The step-by-step explanation of the above question is as follows:

Given:

• The mole fraction of water is 0.85.

To find:

• The molality of the solution?

Formula used:

• Mole fraction: $\frac{n_{1} }{n_{1}+n_{2} }$

Step-by-step explanation:

If $n_{1}$ and $n_{2}$ are the number of the moles of water and $H_{2} SO_{4}$ respectively, then

$x_{water} = \frac{n_{1} }{n_{1} +n_{2} }$ = 0.85 ........................(i)

$x_{H_{2} SO_{4} }$ = $\frac{n_{2} }{n_{1} +n_{2} }$ = 1-0.85 = 0.15 ...................(ii)

Dividing (ii) by (i)

$n_{2}/n_{1}$ = 0.15/0.85

Now $n_{1}$ = 1000/18 = 55.55

∴ $n_{2}$ = 0.15/0.85 x 55.55 = 9.5moles

Hence, molality of solution = 9.8 m.

#SPJ2