Vapour pressure of CCl 4 at 25- C is 143 mm of Hg- 0-5 gm of a non-volatile solute mol- wt- -65 is dissolved in 100 ml CCl 4- Find the vapour pressure of the solution Density of CCl 4-1-58 g - cm 2 A- 199-34 mmB- 141-43 mmC- 143-99 mmD- 94-39 mm

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Relative Lowering of Vapour Pressure

Vapour pressu...

Question

Vapour pressure of CCl₄ at 25⁰C is 143mm of Hg. 0.5 gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl₄. Find the vapour pressure of the solution (Density of CCl₄=1.58g/cm²)

141.43 mm

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94.39 mm

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199.34 mm

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143.99 mm

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Solution

The correct option is
B

141.43 mm
$W_{B} = 0.5 g r a m,$
$M o l e c u l a r w e i g h t o f B (M_{B}) = 65 a m u$
$W_{A} = V o l u m e \times d e n s i t y = 100 \times 1.58 = 158 g r a m s$
$M o l e c u l a r m a s s o f {C C l}_{4} (M_{A}) = 12 + 35.5 \times 4 = 154 g m {o l}^{- 1}$
$\frac{p^{o} - p^{s}}{p^{o}} = \frac{W_{B} \times M_{A}}{M_{B} \times W_{A}} \times p^{o}$
$p^{s} = p^{o} W_{B} \times \frac{M_{A}}{M_{B} \times {W_{A}}} \times p^{o}$
$= 143 - 0.5 \times \frac{154}{65 \times 158} \times 143 = 143 - 1.07 = 141.93 m m$

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Vapour pressure of CCl₄ at 25⁰C is 143mm of Hg. 0.5 gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl₄. Find the vapour pressure of the solution (Density of CCl₄=1.58g/cm²)

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Vapour pressure of CCl4 at 250C is 143mm of Hg. 0.5 gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl4. Find the vapour pressure of the solution (Density of CCl4=1.58g/cm2)

The correct option is B 141.43 mmWB=0.5gram,MolecularweightofB(MB)=65amuWA=Volume×density=100×1.58=158gramsMolecularmassofCCl4(MA)=12+35.5×4=154gmol−1po−pspo=WB×MAMB×WA×pops=poWB×MAMB×WA×po=143−0.5×15465×158×143=143−1.07=141.93mm

Vapour pressure of CCl₄ at 25⁰C is 143mm of Hg. 0.5 gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl₄. Find the vapour pressure of the solution (Density of CCl₄=1.58g/cm²)