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1xx10^(3)m" solution of Pt "(NH(3))(4)Cl...

`1xx10^(3)m" solution of Pt "(NH_(3))_(4)Cl_(4) " in "H_(2)O` shows depression in freezing point by `0.0054^(@)C`. The structure of the compound will be
`("given "K_(f)(H_(2)O)=1.860km^(-1))`

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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) ]

If 0.1 M H_(2)SO_(4) (aq.) solution shows freezing point -0.3906^(@)C then what is the K_(a2)"for"H_(2)SO_(4) ? (Assume m = M and K_(f(H_(2)O) = 1.86 K kg mol^(-1) )

If 0.1 M H_(2)SO_(4) (aq.) solution shows freezing point 0.3906^(@)C then what is the K_(a2)"for"H_(2)SO_(4) ? (Assume m = M and K_(f(H_(2)O) = 1.86 K kg mol^(-1) )

Equal volume of 1.0 M KCl and 1.0 M AgNO_(3) are mixed . The depression of freezing point of the resulting solution will be : ( K_(f) (H_(2)O) = 1.86K kg mol^(-1) , Assume : 1M = 1m)

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A 0.001 molal solution of [Pt(NH_(3))_(4)Cl_(4)] in water had freezing point depression of 0.0054^(@)C . If K_(f) for water is 1.80 , calculating the number f Cl^(-) ions furnished.