A) 1.4
B) 14.0
C) 3.2
D) 2.0
Correct Answer: C
Solution :
[c]| Mole fraction of solute \[{{\chi }_{2}}=0.2.\] |
| Therefore, mole fraction of solvent \[{{\chi }_{1}}=1-{{\chi }_{2}}\] |
| \[\Rightarrow \] \[1-0.2=0.8\] |
| or \[\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=0.2\] and \[\frac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}=0.8\] |
| or \[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{0.2}{0.8}=\frac{1}{4}\] |
| Now, if n1 (solvent moles) \[=\frac{1000}{78}\] |
| \[=12.8\] moles |
| \[{{n}_{2}}=\frac{12.8}{4}=3.2\] moles. Therefore, \[3.2\] moles of the compound are present in one kg of solvent benzene and so molality \[=3.2.\] |
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