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)`)

A 0.025 m solution of mono basic acid had a freezing point of 0.06^(@)C . Calculate K_(a) for the acid K_(f) for H_(2)O=1.86^(@) "molality"^(-1) .

An aqueous solution at freezes at -2.55^(@)C . What is its boiling point (K_(b)^(H_(2)O)=0.52 K m^(-1) , K_(f)^(H_(2)O)=1.86 K m^(-1) ?

An aqueous solution at -2.55^(@)C . What is its boiling point (K_(b)^(H_(2)O)=0.52 K m^(-1) , K_(f)^(H_(2)O)=1.86 K m^(-1) ?

An aqueous solution at -2.55^(@)C . What is its boiling point (K_(b)^(H_(2)O)=0.52 K m^(-1) , K_(f)^(H_(2)O)=1.86 K m^(-1) ?

An aqueous solution of NaCI freezes at -0.186^(@)C . Given that K_(b(H_(2)O)) = 0.512K kg mol^(-1) and K_(f(H_(2)O)) = 1.86K kg mol^(-1) , the elevation in boiling point of this solution is:

An aqueous solution of NaCI freezes at -0.186^(@)C . Given that K_(b(H_(2)O)) = 0.512K kg mol^(-1) and K_(f(H_(2)O)) = 1.86K kg mol^(-1) , the elevation in boiling point of this solution is:

An aqueous solution of NaCI freezes at -0.186^(@)C . Given that K_(b(H_(2)O)) = 0.512K kg mol^(-1) and K_(f(H_(2)O)) = 1.86K kg mol^(-1) , the elevation in boiling point of this solution is:

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) ]