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Irina18 [472]
3 years ago
9

Chloral hydrate (C2H3Cl3O2) is a drug formerly used as a sedative and hypnotic.

Chemistry
1 answer:
Dmitrij [34]3 years ago
3 0

Answer :

(a) The molar mass of C_2H_3Cl_3O_2 is, 165.5 g/mole

(b) The moles of C_2H_3Cl_3O_2 is, 3.02 moles

(c) The mass in grams of 2.0\times 10^{-2} mole chloral hydrate is, 3.31 g

(d) The number of chlorine atoms in 5.0 g chloral hydrate is, 5.4\times 10^{22}

(e) The mass of chloral hydrate will be, 1.55 g

(f) The mass of exactly 500 molecules of chloral hydrate is, 1.99\times 10^{23}

Explanation :

(a) To calculate the molar mass of chloral hydrate.

The formula of chloral hydrate is, C_2H_3Cl_3O_2

Atomic mass of carbon = 12 g/mole

Atomic mass of hydrogen = 1 g/mole

Atomic mass of oxygen = 16 g/mole

Atomic mass of chlorine = 35.5 g/mole

Now we have to determine the molar mass of chloral hydrate.

\text{Molar mass of }C_2H_3Cl_3O_2=2(12g/mole)+3(1g/mole)+3(35.5g/mole)+2(16g/mole)=165.5g/mole[/tex]

The molar mass of C_2H_3Cl_3O_2 is, 165.5 g/mole

(b) Now we have to determine the moles of C_2H_3Cl_3O_2.

\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{500.0g}{165.5g/mole}=3.02moles

The moles of C_2H_3Cl_3O_2 is, 3.02 moles

(c) Now we have to determine the mass in grams of 2.0\times 10^{-2} mole chloral hydrate.

\text{Mass of }C_2H_3Cl_3O_2=\text{Moles of }C_2H_3Cl_3O_2\times \text{Molar mass of }C_2H_3Cl_3O_2

\text{Mass of }C_2H_3Cl_3O_2=(2.0\times 10^{-2}mole)\times (165.5g/mole)=3.31g

The mass in grams of 2.0\times 10^{-2} mole chloral hydrate is, 3.31 g

(d) To calculate the number of chlorine atoms are in 5.0 g chloral hydrate.

First we have to determine the moles of C_2H_3Cl_3O_2.

\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{5g}{165.5g/mole}=0.03moles

Now we have to calculate the number of chlorine atoms in chloral hydrate.

In C_2H_3Cl_3O_2, there are, 2 carbon atoms, 3 hydrogen atoms, 3 chlorine atoms and 2 oxygen atoms.

As, 1 mole of C_2H_3Cl_3O_2 contains 3\times 6.022\times 10^{23} chlorine atoms

So, 0.03 mole of C_2H_3Cl_3O_2 contains 0.03\times 3\times 6.022\times 10^{23}=5.4\times 10^{22} chlorine atoms

The number of chlorine atoms in 5.0 g chloral hydrate is, 5.4\times 10^{22}

(e) To calculate the mass of chloral hydrate would contain 1.0 g Cl.

As, 3\times 35.5g of chlorine present in 165.5 g of C_2H_3Cl_3O_2

So, 1 g of chlorine present in \frac{165.5}{3\times 35.5}=1.55g of C_2H_3Cl_3O_2

The mass of chloral hydrate will be, 1.55 g

(f) To calculate the mass of exactly 500 molecules of chloral hydrate.

As, 6.022\times 10^{23} molecules of chloral hydrate has 165.5 g mass of chloral hydrate

So, 500 molecules of chloral hydrate has \frac{6.022\times 10^{23}}{500}\times 165.5=1.99\times 10^{23} mass of chloral hydrate

The mass of exactly 500 molecules of chloral hydrate is, 1.99\times 10^{23}

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    The equation is  2 Cr  +  3Pb^{2+} \to 3Pb + 2Cr^{3+}

     The  temperature is  T = 25^oC =  298 K   [room  \ temperature ]

     The  emf at standard condition is  E^o_{cell}  =  0.61 \  V

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  =>   G =  6  * 96487 *  0.61

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This Gibbs free energy can also be represented mathematically as

       G =  RTlogK

Here  R  is the cell constant with value 8.314J/K

           K is the equilibrium constant

From above

=>  K  =  antilog^{\frac{G}{ RT} }

Generally  antilog =  2.718

=>K  =  2.718^{\frac{3.5 *10^5}{ 8.314* 298} }

=>   K  =  8*10^{61}

       

         

       

         

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