The mole fraction of a solute in a solution is 0.1 At 298 K, Molarity of this solution is the same as its molality Density of this solution at 298 k is 2.0 g $c m^{3}$ . The ratio of the molecular weights of the solute and solvent,MW solute MW solvent , is?

Open in App

Solution

molecular weight of solute be $= M_{1}$
and molecular weight of solvent be $M_{2}$
fraction of solute $= 0.1$
Let solution be litre
mass of solution $1$ litre
mass of solution $= 2000 g$
mass of solute $= x$ man of solvent $= 2000 - x . g$
$M o l a r i t y = \frac{x}{M_{1}} / 1 l i t r e = \frac{x}{M_{1}}$
$M o l a r i t y = \frac{x}{M_{1}} / (2000 - x) \times 10^{- 3}$
$= \frac{1000 x}{(2000 - x) M_{1}}$
$M o l a r i t y = M o l a r i t y$
$\frac{x}{M_{1}} = \frac{1000 x}{(2000 - x) M_{1}}$
$2000 - x = 1000$
$x = 1000 g M$
number of mole of solute $= \frac{1000}{M_{1}}$
number of mole of solvent $= \frac{1000}{M_{2}}$
$\frac{\frac{1000}{M_{1}}}{\frac{1000}{M_{1}} + \frac{1000}{M_{2}}} \Rightarrow \frac{M_{2}}{M_{1} + M_{2}} = 0.1$
$M_{2} = 0.1 M_{1} + 0.1 M_{2}$
$0.9 M_{2} = 0.1 M_{1} \Rightarrow \frac{M_{1}}{M_{2}} = 9$

Suggest Corrections

Similar questions

0.1

298 K

298 K

2.0 g

{c m}^{- 3}

(\frac{{M W}_{s o l u t e}}{{M W}_{s o l v e n t}})

(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}})

0.1

298 K

298 K

2.0 g c m^{- 1}

(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}})

Join BYJU'S Learning Program