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,

  • m→molality
  • M→molarity
  • Mo→molar mass of solute 
  • d→density

Given that M=0.1M ,d=1.11g/cc ,Mo= 98

  \(\frac{1}{m}=\frac{1.11}{0.1}-\frac{1000}{98}\)

   \(\frac{1}{m}=11.1-10.2 \\ \frac{1}{m}=0.9 \\ m=1.11m\)



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