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(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^0} = x_{	ext{s

Vedclass · Accepted Answer

$A > B > C$

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(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^0} = x_{	ext{solute}} = \frac{n_{	ext{solute}}}{n_{	ext{solute}} + n_{	ext{solvent}}}$.Since the mass of the solvent ($180 \ g$ of water) and the mass of the solutes $(10 \ g)$ are constant,the mole fraction of the solute depends on the number of moles of the solute $(n = \frac{	ext{mass}}{	ext{molar mass}})$.As the molar mass of the solute increases,the number of moles of the solute decreases,which in turn decreases the mole fraction of the solute.Given molar masses: $M_A = 100 \ g \ mol^{-1}$,$M_B = 200 \ g \ mol^{-1}$,$M_C = 10,000 \ g \ mol^{-1}$.Since $M_A  n_B > n_C$.Therefore,the relative lowering of vapour pressure follows the order $A > B > C$.

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The vapour pressure of water at $20\,^{\circ}C$ is $17.5\, mm\, Hg.$ If $18\, g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2\, g$ of water at $20\,^{\circ}C,$ the vapour pressure of the resulting solution will be $.........\, mm$ of $Hg$.

$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
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Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

Relative lowering in vapour pressure of a solution containing a non-volatile solute is the ratio of:

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