1.

6g of a mixture of napthalene (C_10H_8) and anthracene (C_14H_10) is dissolved in 300 gram of benzene. If the depression in freezing point is 0.70 K, the composition of napthalene and anthracene in the mixture respectively in g are (molal depression constant of benzene is 5.1 K mol^-1)

Answer»

2.60,3.40
3.40,2.60
2.90,3.10
3.10,2.90

Solution :GIVEN, weigh of solvent=300g
`Molarity=(mol es of solute)/(weight of solvent(kg))`
`K_J=5.1K kg//mol`
Depression in FREEZING point,
`DELTA T_j=K_j times m`
`0.70=5.1times m`
`0.70/5.1=(Total moles of solute)/(Total weight ofsolvent (kg))`
Let, assume x G napthalene is present,
`0.137=(x/w_(c_10H_8)+(6-x)/W_(c_14H_10) times100)/300`
`0.041=x/(128G)+(6-x)/(178g)implies25x=84.211`
x(napthalene)=34
Now, anthracene =6-x=6-3.4=2.6 g
Hence option (b) is correct.


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