(D) For a dilute solution,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n}{N}$ Where $P^0 - P_s = 0.20 \ mm \ H

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(D) For a dilute solution,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n}{N}$ Where $P^0 - P_s = 0.20 \ mm \ Hg$,$P^0 = 54.2 \ mm \ Hg$,$n$ is the moles of glucose,and $N$ is the moles of water. $N = \frac{100 \ g}{18 \ g \ mol^{-1}} = 5.55 \ mol$. Substituting the values: $\frac{0.20}{54.2} = \frac{w / 180}{100 / 18}$. $\frac{0.20}{54.2} = \frac{w \times 18}{180 \times 100}$. $w = \frac{0.20 \times 180 \times 100}{54.2 \times 18} = \frac{3600}{975.6} \approx 3.69 \ g$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$

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A
$4.69$
B
$3.59$
C
$2.59$
D
$3.69$

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What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$

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$2\, \text{moles}$ of liquid $A$ is mixed with $3\, \text{moles}$ of liquid $B$. Calculate the mole fraction of $A$ in the vapour phase if $P_A^o = 100\, \text{mm}$ and $P_B^o = 200\, \text{mm}$.

If the vapour pressure of a solution containing a non-volatile solute is $2\%$ less than the vapour pressure of pure water,find the molality of the solution.

Vapour pressure of $CCl_{4}$ at $25\,^{\circ} C$ is $143\, mm\, Hg$. $0.5\, g$ of a non-volatile solute (mol. wt. $65$) is dissolved in $100\, mL$ of $CCl_{4}$. Find the vapour pressure of the solution. (Density of $CCl_{4} = 1.58\, g / cm^{3}$)

The ratio of loss in weight of solvent to gain in weight of $CaCl_{2}$ tube in Ostwald-Walker method is:

$12 \ g$ of urea is dissolved in $1 \ L$ of water and $68.4 \ g$ of sucrose is dissolved in $1 \ L$ of water. The lowering of vapour pressure of the first case is

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