The molarity (m) of an aqueous solution containing 22.5 g of glucose (c6h12o6) in 35.5 ml of solution is

Explanation:

Molarity is the number of moles present in a liter of solution.

Mathematically,     Molarity =

And, no. of moles =

Molar mass of is 180.15 g/mol. Therefore, number of moles present will be as follows.

   No. of moles =

                         =

                          = 0.124 mol

Hence, calculate the molarity as follows.

                Molarity =

                       =      (as 1 L = 1000 mL)

                       = 3.51 M

Thus, we can conclude that molarity of the solution is 3.51 M.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"