Question

A solution is prepared by dissolving 50.4 gg sucrose (C12H22O11C12H22O11) in 0.332 kgkg of water. The final volume of the solution is 355 mLmL. For this solution, calculate the molarity.

Answer 1

Answer:

The correct answer is 0.41 M

Explanation:

The molarity of a solution is given by the number of moles of solute per liter of solution.

Molarity= moles solute/1 L solution= mol/L

First we need the molecular weight (Mw) of sucrose in order to convert the mass of sucrose (in g) to mol. We can quickly calculate this from the molar mass (MM) of the elements C, H and O in Periodic Table:

Mw sucrose (C₁₂H₂₂O₁₁)= (12 x MM C) +(22 x MM H) + (11 x MM O)= (12 x 12 g/mol)+(22 x 1 g/mol) + (11 x 16 g/mol)= 342 g/mol

Then, we divide the mass of sucrose (50.4 g) into the molecular weight of sucrose to obtain the number of moles and this is divided into the volume of the solution (355 ml). We convert ml to L taking into account that 1000 ml= 1L:

50.4 g sucrose x 1 mol/342 g sucrose x 1/355 ml x 1000 ml/1 L= 0.415 mol/L= 0.41 M

Answer 2

The molarity of the solution is 0.4146 M.

Based on the given information,

• The mass of sucrose is 50.4 g.

• The final volume of the solution is 355 ml.  

• It is known that the molar mass of sucrose is 342.2965 g/mol.  

Now the molarity of the solution is determined by using the formula,

$Molarity = \frac{Moles of sucrose}{Volume of solution in ml} *1000$

Now the moles of sucrose present in the solution is,

$Moles = \frac{mass of sucrose}{Molar mass of sucrose}$

$Moles = \frac{50.4}{342.2965} \\Moles = 0.1472 moles$

Now putting the values in the formula for molarity we get,

$Molarity = \frac{0.1472}{355} * 1000\\Molarity = 0.4146 M$

Thus, the molarity of the solution is 0.4146 M.

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