Question

A solution is prepared by dissolving 318.6 g sucrose (C12H22O11) in 4905 g of water. Determine the molarity of the solution

Answer 1

The molarity of sucrose solution is 0.19 M.

The molarity of HCl is 12.8 M.

Explanation:

a. Molarity can be found by finding its moles and volume of water in L and then dividing both(moles divided by volume in Litres).

Mass of sucrose = 318. 6 g

Molar mass of sucrose = 342.3 g/mol

Moles = $\frac{mass}{molar mass}$

       = $\frac{318.6 g}{342.3 g/mol}$

      = 0.93 moles

Mass of water = 4905 g

Density of water = 1000 g/L

Volume = $\frac{mass}{density}$

             = $\frac{4905 g }{1000 g/L}$

          = 4.905 L

Now we can find the molarity = $\frac{moles}{Volume(L)}$

                               =  $\frac{0.93 moles}{4.905 L}$

                              = 0.19 M

So the molarity of sucrose solution is 0.19 M.

b. The molarity of HCl can be found as follows.

It is given that 39% HCl that means it contains 39 g of acid in 100 g of water.

Density of the solution is 1.20 g/mL, from this mass can be found as,

$\frac{1 L \times 1000 mL \times 1.20 g}{1 L \times 1 mL}$

= 1200 g

Now we have to find out the amount of HCl in grams as,

$\frac{1200g \times 39 g HCl }{100 g solution}$

= 468 g HCl

Now we have to find the number of moles,

moles = $\frac{468 g}{36.46 g/mol}$

           = 12.8 moles

Molarity of HCl = $\frac{12.8 moles}{1 L }$

                          = 12.8 M

So the molarity of HCl is 12.8 M.