If 1 g of solute dimerise upto `75%` in 100 g of `H_(2)O` and depression in freezing point `0.093^(@)` C. Find average molecular weight of solute after dimerisation `(K_(f(H_(2)O))=1.86)`.

6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C . The molecular mass of the substance will be: ( K_(f) for water = 1.86^(@)C /molal)

6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C . The molecular mass of the substance will be: ( K_(f) for water = 1.86^(@)C /molal)

When 45 g of an unknown compound was dissolved in 500 g of water, the solution has freezing point of -0.93^(@)C (i) What is the molecular weight of compound ? (K_(f)=1.86) (ii) If empirical formula is CH_(2)O , what is the molecular formula of compound ?

When 45 g of an unknown compound was dissolved in 500 g of water, the solution has freezing point of -0.93^(@)C (i) What is the molecular weight of compound ? (K_(f)=1.86) (ii) If empirical formula is CH_(2)O , what is the molecular formula of compound ?

When 45 g of an unknown compound was dissolved in 500 g of water, the solution has freezing point of -0.93^(@)C (i) What is the molecular weight of compound ? (K_(f)=1.86) (ii) If empirical formula is CH_(2)O , what is the molecular formula of compound ?

1 xx 10^(-3) m solution of Pt(NH_(3))_(4)Cl_(4) in H_(2)O shows depression in freezing point of 0.0054^(@)C . The formula of the compound will be [Given K_(f) (H_(2)O) = 1.86^(@)C m^(-1) ]

1 xx 10^(-3) m solution of Pt(NH_(3))_(4)Cl_(4) in H_(2)O shows depression in freezing point of 0.0054^(@)C . The formula of the compound will be [Given K_(f) (H_(2)O) = 1.86^(@)C m^(-1) ]

75.2 g of C_(6)H_(5)OH (phenol) is dissolved in a solvent of K_(f) = 14 . If the depression in freezing point is 7K , then find the percentage of phenol that dimerises.