Question

The hydrochloride form of cocaine has a solubility of 1.00 g in 0.400 mL water. Calculate the molarity of a saturated solution of the hydrochloride form of cocaine in water.Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

The molarity of the solution is 7.4 mol/L

Explanation:

From the question above

0.400 ml of water contains 1.00 g of hydrochloride form of cocaine

Therefore 1000 ml of water will contain x g of hydrochloride form of cocaine

                    x = 1000 / 0.400

                    x = 2500 g

2500g of hydrochloride form of cocaine is present in 1000 ml of water.

Mole of hydrochloride form of cocaine = mass /molar mass of hydrochloride

Mole of hydrochloride form of cocaine = 2500/339.8

                                                                = 7.4 mol

Molarity = mol/ volume in liter (L)

molarity = 7.4 /1

Molarity = 7.4 mol/L

Answer 2

Answer:

$M=7.35M$

Explanation:

Hello,

In this case, the cocaine hydrochloride whose molecular formula is:

$C_{17}H_{22}ClNO_4$

Has a molar mass of 339.8 g/mol, for that reason, in 1.00 g there are the following moles:

$n=1.00gC_{17}H_{22}ClNO_4*\frac{1molC_{17}H_{22}ClNO_4}{339.8gC_{17}H_{22}ClNO_4} =2.94x10^{-3}molC_{17}H_{22}ClNO_4$

Such calculations are performed at the saturation condition with which the molarity is obtained as:

$M=\frac{n}{V}$

Thus, the volume in liters is:

$V=0.4mL*\frac{1L}{1000mL} =4x10^{-4}L$

As we assume the volume does not change when the cocaine hydrochloride is added to the water, therefore, we obtain the molarity:

$M=\frac{2.94x10^{-3}molC_{17}H_{22}ClNO_4}{4.00x10^{-4}L} =7.35M$

Best regards.