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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: 0.5g KCl was dissolved in 100g water and the solution originally .........
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sneha.bagri (142)

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 0.5g KCl was dissolved in 100g water and the solution originally at 200C, froze at  
-0.240C. Calculate the percentage ionization of salt. Kf per 1000g of water = 1.86K.
 
    
rockey (168)

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delta Tf =Kf molality

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sneha.bagri (142)

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here we need to consider vant hoff factor me not getting d ans....
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sneha.bagri (142)

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hey frnds dis is cbse modle paper ques so plz try dis.........
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avi_1214545 (974)

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is it .92?

Na pine ka shouk tha na pilane ka shonk tha. Hume to sirf NAZREIN milane ka shonk tha. Par kya karen hum nazre hi unse mila baithe jinhe nazron se PILANE ka shonk tha.
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dheeraj89 (0)

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^T= imK
.24= [1+(n-1)@] * .067 * 1.86  
[Here n=2 bcoz 2 ions & @= degree of ionisation]
after solving,
@=.9225  or 92.25%
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ashish_banga (1016)

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delta Tf = i Kf X m
m = 0.5 X 10 / 74.5
get i and use 1 + x = i
x = degree of dissociation
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