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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

The vapour pr...

Question

The vapour pressure of pure benzene is $640 m m$ of $H g$ . $2.175 \times 10^{- 3} k g$ of non-volatile solute is added to $39$ gram of benzene, the vapour pressure of solution is $600 m m$ of $H g$ . Calculate molar mass of solute $(C - 12, H = 1)$ .

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Solution

Molar mass of benzene $= (6 \times 12 + 6 \times 1)$
$= 78 \times 10^{- 3} k g {m o l}^{- 1}$

$P^{o} = 640 m m$ $H g$ ; $P = 600 m m$ $H g$

$W_{1} = 39 \times 10^{- 3} k g$ , $W_{2} = 2.175 \times 10^{- 3} k g$

$M_{1} = 78 \times 10^{- 3} k g$ , $M_{2} =$ ?

$M_{2} = \frac{P^{o} \times W_{2} \times M_{1}}{Δ P \times W_{1}}$

$M_{2} = \frac{640 \times 2.175 \times 10^{- 3} \times 78 \times 10^{- 3}}{(640 - 600) \times 39 \times 10^{- 3}}$

$= 69.6 \times 10^{- 3} k g {m o l}^{- 1}$

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The vapour pressure of pure benzene is 640mm of Hg. 2.175×10−3kg of non-volatile solute is added to 39 gram of benzene, the vapour pressure of solution is 600mm of Hg. Calculate molar mass of solute (C−12,H=1).

Molar mass of benzene =(6×12+6×1) =78×10−3kgmol−1Po=640mm Hg; P=600mm HgW1=39×10−3kg, W2=2.175×10−3kgM1=78×10−3kg, M2=?M2=Po×W2×M1ΔP×W1M2=640×2.175×10−3×78×10−3(640−600)×39×10−3 =69.6×10−3kgmol−1

The vapour pressure of pure benzene is $640 m m$ of $H g$ . $2.175 \times 10^{- 3} k g$ of non-volatile solute is added to $39$ gram of benzene, the vapour pressure of solution is $600 m m$ of $H g$ . Calculate molar mass of solute $(C - 12, H = 1)$ .