Question

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

Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molarity of the HCl, if the solution has a density of 1.20 g/mL.


Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the mol of the HCl, if the solution has a density of 1.20 g/mL.

Answer 1

The molarity of sucrose solution is 0.19 M.

The molarity of HCl is 12.8 M.

Step-by-step 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 =
`$(mass)/(molar mass)`

=
`$(318.6 g)/(342.3 g/mol)`

= 0.93 moles

Mass of water = 4905 g

Density of water = 1000 g/L

Volume =
`$(mass)/(density)`

=
`$(4905 g )/(1000 g/L)`

= 4.905 L

Now we can find the molarity =
`$(moles)/(Volume(L))`

=
`$(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,

`$(1 L * 1000 mL * 1.20 g)/(1 L * 1 mL)`

= 1200 g

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

`$(1200g * 39 g HCl )/(100 g solution)`

= 468 g HCl

Now we have to find the number of moles,

moles =
`$(468 g)/(36.46 g/mol)`

= 12.8 moles

Molarity of HCl =
`$(12.8 moles)/(1 L )`

= 12.8 M

So the molarity of HCl is 12.8 M.