The mole fraction of a solute in a 100 molal aqueous solution is -10-2- Round off to the nearest integer- -Given - Atomic mass of H - 1-0 u - Atomic mass of O - 16-0 u-

Let weight of nbsp; H_2O = 1000 g Moles of solute = 100 Moles of nbsp; H_2O = 100018 Moles fraction of solute =mole of soluteTotal moles=100100+ 100018 ...

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Standard XII

Chemistry

Solubility Product

The mole frac...

Question

The mole fraction of a solute in a 100 molal aqueous solution is $\times 10^{- 2}$ . (Round off to the nearest integer). [Given : Atomic mass of H = 1.0 u ; Atomic mass of O = 16.0 u]

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Solution

Let weight of $H_{2} O = 1000 g$

Moles of solute = 100

Moles of $H_{2} O = \frac{1000}{18}$

$Moles fraction of solute =moleofsoluteTotalmoles=100100+100018$
$Moles fraction of solute =18002800$
$X_{s o l u t e} \approx 64 \times 10^{- 2}$

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The mole fraction of a solute in a 100 molal aqueous solution is ×10−2. (Round off to the nearest integer). [Given : Atomic mass of H = 1.0 u ; Atomic mass of O = 16.0 u]

Let weight of H2O=1000g Moles of solute = 100 Moles of H2O=100018 Moles fraction of solute =moleofsoluteTotalmoles=100100+100018 Moles fraction of solute =18002800 Xsolute≈64×10−2

The mole fraction of a solute in a 100 molal aqueous solution is $\times 10^{- 2}$ . (Round off to the nearest integer). [Given : Atomic mass of H = 1.0 u ; Atomic mass of O = 16.0 u]

Let weight of $H_{2} O = 1000 g$

Moles of solute = 100

Moles of $H_{2} O = \frac{1000}{18}$

$Moles fraction of solute =moleofsoluteTotalmoles=100100+100018$
$Moles fraction of solute =18002800$
$X_{s o l u t e} \approx 64 \times 10^{- 2}$