$1000 g$ an aqueous solution contains $100 g$ of a solute X of molar mass 200. The mole fraction of solute in the solution is:

0.2

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0.009

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0.25

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0.125

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Solution

The correct option is A $0.009$
$x = \frac{\frac{100}{200}}{\frac{100}{200} + \frac{1000}{18}} = \frac{0.5}{0.5 + 55.5}$

$= \frac{0.5}{56} = 0.009$

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1000g an aqueous solution contains 100g of a solute X of molar mass 200. The mole fraction of solute in the solution is:

The correct option is A 0.009x=100200100200+100018=0.50.5+55.5 =0.556=0.009

$1000 g$ an aqueous solution contains $100 g$ of a solute X of molar mass 200. The mole fraction of solute in the solution is:

The correct option is A $0.009$
$x = \frac{\frac{100}{200}}{\frac{100}{200} + \frac{1000}{18}} = \frac{0.5}{0.5 + 55.5}$

$= \frac{0.5}{56} = 0.009$