30 mL of 0.1M KI(aq.) and 10 mL of 0.2M ...

`30 mL` of `0.1M KI_(aq.)` and `10 mL` of `0.2M AgNO_(3)` are mixed. The solution is then filtered out. Assuming that no change in total volume, the resulting solution will freezing at:
`[K_(f) "for" H_(2)O = 1.86K kg mol^(-1)`, assume molality = molality]

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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)

0.1 M KI and 0.2 M AgNO_(3) are mixed in 3 : 1 volume ratio. The depression of freezing point of the resulting solution will be [ K_(b)(H_(2)O) = 1.86 K kg "mol"^(-1) ]:

0.1M KI and 0.2 M AgNO_(3) are mixed in 3:1 volume ratio . The deperession in freezing point of the resulting solution will be 0.1 x (Assume Kf of H_(2)O=2Kg//"mol" ).

0.1 M KI and 0.2 M AgNO_(3) are mixed in 3 : 1 volume ratio. The depression of freezing point of the resulting solution will be [ K_(b)(H_(2)O) = 1.86 K kg "mol"^(-1) ]: (a)3.72 K (b)1.86 K (c)0.93 K (d)0.279 K

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

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

`30 mL` of `0.1M KI_(aq.)` and `10 mL` of `0.2M AgNO_(3)` are mixed. The solution is then filtered out. Assuming that no change in total volume, the resulting solution will freezing at: `[K_(f) "for" H_(2)O = 1.86K kg mol^(-1)`, assume molality = molality]

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`30 mL` of `0.1M KI_(aq.)` and `10 mL` of `0.2M AgNO_(3)` are mixed. The solution is then filtered out. Assuming that no change in total volume, the resulting solution will freezing at:
`[K_(f) "for" H_(2)O = 1.86K kg mol^(-1)`, assume molality = molality]