Question

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Calculate the molarity of the solutions described below.
a. 10.0 g of sodium chloride is dissolved in 2.0 L of solution.
b. 52.5 g of sugar (C12H22O11) s dissolved in 1.0 L of solution.
c. 120 g of aluminum sulfate is dissolved in 10.0 L of solution.
d. 1.75 g of caffeine (C8H10N4O2) is dissolved in 0.100 L of solution.

Answer 1

a. 0.0855 M

b. 0.153 M

c. 0.0351 M

d. 0.0901 M

Step-by-step explanation:

a.

Molar mass NaCL= 23.0 +35.5 = 58.5 g/mol

10.0 g* 1 mol/58.5 g = 10.0/58.5 mol

(10.0/58.5) mol/2L = 0.0855 mol/L = 0.0855 M

b.

Molar mass (C12H22O11) =342.3 g/mol

52.5 g*1mol/342.3 g = 52.5/342.3 mol=0.153 mol

0.153 mol/1L = 0.153 M

c.

Molar mass (Al2(SO4)3) =342.2 g/mol

120g* 1 mol/342.2 g =0.351 mol

0.351 mol/10.0L = 0.0351 mol/L = 0.0351 M

d.

Molar mass (C8H10N4O2) = 194.2 g/mol

1.75 g*1mol/194.2 g = 0.00901 mol

0.00901 mol/0.100 L =0.0901 M