1.

Calculate the mass of ascorbic acid (Molar mass = 176 g mol^(-1)) to be dissolved in 75 g of acetic acid, to lower its freezing point by 1.5^(@)C. [K_(f)=3.9"K kg mol"^(-1)]

Answer»

Solution :Apply the relation
`M_(2)=(K_(f)XX w_(2) xx 1000)/(DeltaT_(f) xx w_(1))`
Given: `K_(f)=3.9"K kg MOL"^(-1), M_(2)=176" g mol"^(-1), DeltaT_(f)=1.5""^(@)C, w_(1)=75g`
Substituting the values in the equation above, we have
`176" g mol"^(-1)=(3.9"K kg mol"^(-1) xx w_(2) xx 1000)/(1.5K xx 75g)`
or `w_(2)=(176"g mol"^(-1)xx1.5K xx 75g)/(3.9"K kg mol"^(-1) xx 1000"g kg"^(-1))`
`=(176xx1.5xx75)/(3.9xx1000)=5.077g`


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