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2 years ago

The density of a 45.0 mass % solution of ethanol (C2H5OH) in water is 0.873 g/mL. What is the molarity of the solution?

Chemistry

1 answer:

antoniya [11.8K]2 years ago

6 0

Answer : The concentration of solution is, 8.53 M.

Explanation :

As we are given, 45.0 mass % solution of ethanol in water that means 45.0 g of ethanol present in 100 g of solution.

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}$

$\text{Volume of solution}=\frac{100g}{0.873g/mL}=114.5mL$

Now we have to calculate the molarity of solution.

Mass of $C_2H_5OH$ = 45.0 g

Volume of solution = 114.5 mL

Molar mass of $C_2H_5OH$ = 46.07 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }C_2H_5OH\times 1000}{\text{Molar mass of }C_2H_5OH\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{45.0g\times 1000}{46.07g/mole\times 114.5mL}=8.53mole/L=8.53M$

Therefore, the concentration of solution is, 8.53 M.

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The molecular formula of cisplatin is $Pt(NH_3)_2Cl_2$ .

The atomic weight of the elements in cisplatin is:

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Nitrogen, $N = 14.0067 u$

Hydrogen, $H = 1 u$

Chlorine, $Cl = 35.453 u$

The molar mass of $Pt(NH_3)_2Cl_2$ = $195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453)$ = $300.00 g/mol$

The mass of each element calculated using formula (1):

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$\frac{195.084}{300.00} \times 100 = 65.23$ %.

- Nitrogen, $N$ %

$\frac{2\times 14.0067}{300.00} \times 100 = 9.34$ %

- Hydrogen, $H$ %

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$\frac{2\times 35.453}{300.00} \times 100 = 23.63$ %

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$K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)$

According to the balanced reaction, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ .

Now, calculating the number of moles of $K_2PtCl_4$ in 100.0 g.

Number of moles = $\frac{given mass}{Molar mass}$

Molar mass of $K_2PtCl_4$ = $2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol$

Number of moles of $K_2PtCl_4$ = $\frac{100 g}{415.093 g/mol} = 0.241 mole$ .

Since, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ . Therefore, mass of cisplatin is:

$0.241 mole\times 300 g/mol = 72.3 g$

For mass of $KCl$ :

Molar mass of $KCl$ = $39.0983 + 35.453 = 74.55 g/mol$

Since, 1 mole of $K_2PtCl_4$ gives 2 mole of $KCl$ . Therefore, mass of $KCl$ is:

$0.241 mole\times 74.55 g/mol\times 2 = 35.93 g$

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