The mole fraction of a solute in a 100 molal aqueous solution.× 10-2 (Round off to the Nearest Integer). [Given: Atomic masses: H: 1.0 u , O: 16.0 u]

Question

The mole fraction of a solute in a 100 molal aqueous solution.× 10-2 (Round off to the Nearest Integer). [Given: Atomic masses: H: 1.0 u , O: 16.0 u] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

100 molal aqueous solution means there is 100 mole solute in 1 kg =1000 gm water. Now, mole-fraction of solute =( n text solute / n text solute + n text solvent ) =(100/100+(1000)18)=(1800/2800)=0.6428 =64.28 × 10-2

Q. The mole fraction of a solute in a $100 m o l a l$ aqueous solution_________. $\times 1 0^{- 2}$ (Round off to the Nearest Integer).
[Given : Atomic masses : $H : 1.0 u, O : 16.0 u]$

$100$ molal aqueous solution means there is $100$ mole

solute in $1 k g = 1000 g m$ water.

Now,

mole-fraction of solute $= \frac{n _{solute}}{n _{solute} + n _{solvent}}$

$= \frac{100}{100 + \frac{1000}{18}} = \frac{1800}{2800} = 0.6428$

$= 64.28 \times 1 0^{- 2}$

Q. The mole fraction of a solute in a 100molal aqueous solution_________.×10−2 (Round off to the Nearest Integer). [Given : Atomic masses : H:1.0u,O:16.0u]

100 molal aqueous solution means there is 100 mole solute in 1kg=1000gm water. Now, mole-fraction of solute =nsolute ​+nsolvent ​nsolute ​​ =100+181000​100​=28001800​=0.6428 =64.28×10−2

Q. The mole fraction of a solute in a $100 m o l a l$ aqueous solution_________. $\times 1 0^{- 2}$ (Round off to the Nearest Integer).
[Given : Atomic masses : $H : 1.0 u, O : 16.0 u]$

$100$ molal aqueous solution means there is $100$ mole

solute in $1 k g = 1000 g m$ water.

Now,

mole-fraction of solute $= \frac{n _{solute}}{n _{solute} + n _{solvent}}$

$= \frac{100}{100 + \frac{1000}{18}} = \frac{1800}{2800} = 0.6428$

$= 64.28 \times 1 0^{- 2}$