the vapour pressure of benzene at 80 o C is lowered by 10 mm by dissolving 2 g of a non- volatile substance in 78 g of benzene . the vapour pressure o f pure benzene at 80 o C is 750 mm .the molecular mass of the substance will be____

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Bhavya · Answer

Dear student relative lowering of V.P (p o -ps/po) = mole fraction of solute 750-740/750 = (2/M)/ 2/M + 78/78 10/750 = (2/M) / 2/M + 1 M = 148

Jms · Answer

po-ps/po=w2/m2 × m1/w1 750-740/750=2/m2×78/78 10/750=2/m2 M2 =150.please uovote it if it is right..(question is from chapter solutions standard 12 )

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the vapour pressure of benzene at 80 o C is lowered by 10 mm by dissolving 2 g of a non- volatile substance in 78 g of benzene . the vapour pressure o f pure benzene at 80 o C is 750 mm .the molecular mass of the substance will be____

the vapour pressure of benzene at 80 ^oC is lowered by 10 mm by dissolving 2 g of a non- volatile substance in 78 g of benzene . the vapour pressure o f pure benzene at 80^oC is 750 mm .the molecular mass of the substance will be____

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the vapour pressure of benzene at 80 o C is lowered by 10 mm by dissolving 2 g of a non- volatile substance in 78 g of benzene . the vapour pressure o f pure benzene at 80 o C is 750 mm .the molecular mass of the substance will be____

the vapour pressure of benzene at 80 oC is lowered by 10 mm by dissolving 2 g of a non- volatile substance in 78 g of benzene . the vapour pressure o f pure benzene at 80oC is 750 mm .the molecular mass of the substance will be____

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the vapour pressure of benzene at 80 ^oC is lowered by 10 mm by dissolving 2 g of a non- volatile substance in 78 g of benzene . the vapour pressure o f pure benzene at 80^oC is 750 mm .the molecular mass of the substance will be____