Question

A student prepares a solution by dissolving 60.00 g of glucose (molar mass 180.2 g mol-1) in enough distilled water to make 250.0 mL of solution. The molarity of the solution should be reported as
a. 12.01 M
b. 12.0 M
c. 1.332 M
d. 1.33 M
e. 1.3 M

Answer 1

Answer:

1.332 Molar is the molarity of the glucose solution.

Explanation:

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of glucose = 60.00 g

Molar mass of glucose = 180.2 g/mol

Volume of the solution = V = 250.0 mL = 0.250 L

(1 mL  = 0.001 L)

$Molarity=\frac{60.00 g}{180.2 g/mol\times 0.250 L}=1.332 mol/L$

1.332 Molar is the molarity of the glucose solution.