In the determination of molecular mass by Victor - Meyer's Method 0.790 g of a volatile liquid displaced 1.696 xx 10^(-4) m^(3) of moist air at 303 K and at1 xx 10^(5) Nm^(-2) pressure. Aqueous tension at 303 K is4.242 xx 10^(3) Nm^(-2) . Calculate the molecular mass and vapour density of the compound .

In the determination of molecular mass by Victor - Meyer's Method 0.79

1.	In the determination of molecular mass by Victor - Meyer's Method 0.790 g of a volatile liquid displaced 1.696 xx 10^(-4) m^(3) of moist air at 303 K and at1 xx 10^(5) Nm^(-2) pressure. Aqueous tension at 303 K is4.242 xx 10^(3) Nm^(-2) . Calculate the molecular mass and vapour density of the compound .
Answer» Solution :Mass of the ORGANIC compound =0.79 g Volume of VAPOUR = `V_1 = 1.696 xx 10^(-4) m^(3)` Volume of air displaced = Volume of vapour. `P_1 `= ( atmospheric pressure - aqueous tension ) `=(1.0 xx 10^5) - (4.242 xx 10^(3)) = 0.958 xx 10^(5) Nm^(-2)` `T_1 `= 303 K `(P_1 V_1)/( T_1) = (P_0 V_0)/( T_0)` `V_0 = (P_1 V_1 T_0)/(P_0 T_1)` `V_0 = (0.958 xx 10^(5) xx 1.696 xx 10^(-4))/(1.013 xx 10^(3)) xx (273)/(303)` `V_0 = 1.445 xx10^(-4) m^3` The mass of `1.445 xx 10^(-4)m^(3)` of vapour at S.T.P= 0.79 g . The mass of ` 2.24 xx 10^(-2) m^(3)` of vapour of S.T.P is `= (2.24 xx 10^(-2) xx 0.79)/( 1.445 xx 10^(-4))` The molecular mass of the substance = 122 .46 Vapour density of the compound `=(" molecular mass ")/(2)` `= (122.46)/(2) = 61.23`