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2 years ago

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The hydrochloride form of cocaine has a solubility of 1.00 g in 0.400 mL water. Calculate the molarity of a saturated solution o

f the hydrochloride form of cocaine in water.Express your answer to three significant figures and include the appropriate units.

2 answers:

Tanya [424]2 years ago

6 0

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

TEA [102]2 years ago

4 0

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.

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Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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2 years ago