Answer :
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 = [tex]$\frac{mass}{molar mass}[/tex]
= [tex]$\frac{318.6 g}{342.3 g/mol}[/tex]
= 0.93 moles
Mass of water = 4905 g
Density of water = 1000 g/L
Volume = [tex]$\frac{mass}{density}[/tex]
= [tex]$\frac{4905 g }{1000 g/L}[/tex]
= 4.905 L
Now we can find the molarity = [tex]$\frac{moles}{Volume(L)}[/tex]
= [tex]$\frac{0.93 moles}{4.905 L}[/tex]
= 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,
[tex]$\frac{1 L \times 1000 mL \times 1.20 g}{1 L \times 1 mL}[/tex]
= 1200 g
Now we have to find out the amount of HCl in grams as,
[tex]$\frac{1200g \times 39 g HCl }{100 g solution}[/tex]
= 468 g HCl
Now we have to find the number of moles,
moles = [tex]$\frac{468 g}{36.46 g/mol}[/tex]
= 12.8 moles
Molarity of HCl = [tex]$\frac{12.8 moles}{1 L }[/tex]
= 12.8 M
So the molarity of HCl is 12.8 M.