What are the molality and mole fraction of solute in a 22.2 percent by mass aqueous solution of formic acid (hcooh)?

Calculate  the  mass  of  water  used
that  is
100-22.2=77.8g  convert   into  Kg = 77.8/1000=0.0778Kg  of  water

then  calculate  the  moles  of HCOOH  used
that   is  22.2g/molar  mass  of HCOOH(1+12+16+16+1)=46
therefore  the  moles  of HCOOH=22.2/46=0.48moles
 the  mole  of  water=  77.8/18(molar  mass  of  water= 4.32moles

the   molarity of  HCOOH =  0.48mol/0.0778kg=6.17M

The   mole  ratio= moles  of HCOOH  divided  by  total  moles
the  total  moles=  0.48+4.32=4.8moles
therefore  the  mole  ratio=   0.48/4.8moles=0.1(the  moles  fraction  of  HCOOH)

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Answer charle's law