The relative lowering of vapour pressure of a dilute aqueous solution containig non volatile solute is 0.012. The molality of solution is about

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relative lowering of vap pressure for non volatile solute :
(p⁰ - p_s ) /p⁰= x₂= n₂ / n₁
putting n₁= 1000/18 moles
we get
n₂= 0.012* 1000/18 =0.6667 moles
and this gives molality of solution i.e 0.6667 m

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The relative lowering of vapour pressure of a dilute aqueous solution containig non volatile solute is 0.012. The molality of solution is about

relative lowering of vap pressure for non volatile solute :(p0 - ps ) /p0 = x2 = n2 / n1putting n1 = 1000/18 moles we get n2 = 0.012* 1000/18 =0.6667 moles and this gives molality of solution i.e 0.6667 m

relative lowering of vap pressure for non volatile solute :
(p⁰ - p_s ) /p⁰= x₂= n₂ / n₁
putting n₁= 1000/18 moles
we get
n₂= 0.012* 1000/18 =0.6667 moles
and this gives molality of solution i.e 0.6667 m