- 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
- The vapour pressure of pure benzene at a certain temperature is 640 mm
- The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non
- The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non volatile and non electrolyte solid
- The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non volatile non electrolyte solid weighing 2.175 g is added to 39 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance?
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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
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 × d e n s i t y = 100 × 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 × 4 = 154 g m o l − 1 p o − p s p o = W B × M A M B × W A × p o p s = p o W B × M A M B × W A × p o = 143 − 0.5 × 154 65 × 158 × 143 = 143 − 1.07 = 141.93 m m
The vapour pressure of pure benzene at a certain temperature is 640 mm
The correct Answer is 65.25 According to Raoult.s law P 0 − P P ∘ = w / m w / m + W / M Here, P ∘ = 640 m m H g , P = 600 m m H g , w = 2.175 g W= 39.0g M=78, m = Molecular weight of solute Substituting the various values in the above equation = 640 − 600 640 = 2.175 / m 2.175 / m + 39 / 78 m = 65. 25 g
The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non
The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte solid, weighing 2.175 g is added to 39.08 of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of solid substance? (a) 69.5 (b) 59.6 (c) 49.50 (d) 79.8
The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non volatile and non electrolyte solid
The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non volatile and non electrolyte solid weighing 2.175 g is added to 39.08 of benzene. If the vapour pressure of the solution is 600mm of Hg, what is the molecular weight of solid substance? (a) 49.50 (b) 59.60 (c) 69.60 (d) 79.87
The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non volatile non electrolyte solid weighing 2.175 g is added to 39 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance?
The correct option is B 65.25 Given that Weight of solute = 2.175 g Weight of solvent = 39 g Molar mass of solvent = 78 g Let the molar mass of solute=M n = 2.175 M ; N = 39 78 n is mole of solute and N is mole of solvent p 0 = 640 mm Hg , p = 600 mm Hg (M is the mol. wt. of solute) We know, p 0 − p p 0 = n n + N ∴ 640 − 600 640 = 2.175 / M 2.175 M + 0.5 M = 65.25 g Hence option
