If 0.1 M `H_(2)SO_(4)`(aq.) solution showa 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) )

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) ) (a)0.122 (b)0.0122 (d)1.11x 10^(-3) (d)None of these

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 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: