Saved Bookmarks
| 1. |
The density of \( 0.1 M \) solution of sulphuric acid is \( 1.11 g / cc \). Estimate its molality. |
|
Answer» By relation, \(\frac{1}{m}=\frac{d}{M}-\frac{1000}{M_o}\) Where,
Given that M=0.1M ,d=1.11g/cc ,Mo= 98 \(\frac{1}{m}=\frac{1.11}{0.1}-\frac{1000}{98}\) \(\frac{1}{m}=11.1-10.2 \\ \frac{1}{m}=0.9 \\ m=1.11m\) |
|