Suppose you start with 1.80 g of mgco3·5h2o, after complete dehydration how many grams of the sample do you have remaining?

The  number of  grams of the sample that remained  after  hydration is 0.87  grams

                calculation
calculate the molar  mass of MgCO3. 5 H2O  which is

24 + 12+ (3 x16) + ( 5  x18)  =  174 g/mol

we  well  know  after  hydration  water evaporate leaving behind   MgCO3
therefore the  mass  remained  is  for  MgCO3

find  the   molar  mass of  MgCO3 =  24 +12 +( 3 x16)  = 84  g/mol

the  mass  remained is therefore

 =  molar  mass of MgCO3 /  molar mass of MgCO3.5H2O  x 1.80 g

=  84g/mol / 174 g/mol  x1.8g = 0.87  grams

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)