A) 0.177
B) 1.770
C) 0.0354
D) 0.0177
Correct Answer: D
Solution :
| Molality(m) \[=\frac{1000\times n}{N\times M}\] |
| where, n = number of moles of solute |
| N = number of moles of solvent |
| M = molar mass of solvent |
| Given, \[m=1\] |
| \[\therefore \] \[1=\frac{1000\times n}{N\times 18}\Rightarrow \frac{n}{N}-=\frac{18}{1000}\] |
| or \[\frac{n}{n+N}=\frac{18}{1018}=0.0177\] |
| Alternate Method \[N=\frac{1000}{18}=55.5mol\] |
| \[n=1\] |
| [\[\because \] 1 m solution implies that 1 mole of |
| solute is present in 1 kg or 1000 g water] |
| \[\therefore \] Mole fraction of solute |
| \[=\frac{n}{n+N}=\frac{1}{1+55.5}=0.0177\] |
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