Question

Determine the molarity of a solution prepared by dissolving 141.6 g of citric acid, C3H5O(COOH)3, in 3500.0 mL of water ?

Answer 1

0.21M of C3H5O(COOH)3

Step-by-step explanation:

Molarity = no of mole/ volume of solution

Molar mass of C3H5O (COOH)3 = 192.12g

No of mole of C3H5O (COOH)3 = 141.6/192.12 = 0.74mol

1000mL = 1L

3500mL = 3.5L

Molarity = 0.74/ 3.5 = 0.21M of C3H5O (COOH)3

Answer 2

0.2M

Step-by-step explanation:

The molarity of a solution is given by;

M = number of moles / volume

Number of moles = mass/molar mass

Mass of Citric acid = 141.6g

Molar mass of citric acid = (6×12)+(8×1)+(7×16)

= 72+8+112=192g/mol

Number of moles of citric acid=141.6/192=0.74mol

Volume = 3500.0 mL = (3500/1000)L=3.5L

Molarity of the solution = 0.74mol/3.5L=0.2mol/L

Therefore, 0.2mol/L or 0.2M concentration is prepared by dissolving 141.6 g of citric acid, C3H5O(COOH)3, in 3500.0 mL of water