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kykrilka [37]
2 years ago
15

What mass of CCl4 is required to prepare a 0.25 m solution using 115 g of hexane? (molar mass of ccl4 = 153.81 g/mol and molar m

ass of hexane = 86.17 g/mol.)
Chemistry
1 answer:
natka813 [3]2 years ago
7 0

4.42 g mass of CCl4 is required to prepare a 0.25 m solution using 115 g of hexane.

It's easy to find the molecular mass of a compound with these steps: Determine the molecular formula of the molecule. Use the periodic table to determine the atomic mass of each element in the molecule. Multiply each element's atomic mass by the number of atoms of that element in the molecule.

The molar mass of any compound can be found out by adding the relative atomic masses of each element present in that particular compound.

Hexane is an organic compound, a straight-chain alkane with six carbon atoms and has the molecular formula C₆H₁₄.

Therefore,

⇒ 0.115 g of Hexane x (0.25 mol CCl4/1 mol hexane) x (153.81 g of CCl4/1 mol CCl4) = 4.42g CCl4.

To learn more about CCL4 and Hexane here

brainly.com/question/15156642

#SPJ4

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If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution
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Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

\epsilon = molar absorptivity of this solution =5.56\times 10^4 M^{-1} cm^{-1}

0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c

c=1.4163\times 10^{-5} M=14.16 \mu M

(1\mu M=10^{-6} M)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) c=1.4163\times 10^{-5} mol/L

1 L of solution contains 1.4163\times 10^{-5} moles of red dye.

Mass of 1.4163\times 10^{-5} moles of red dye:

1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g

(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100

red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = c=1.4163\times 10^{-5} M

Concentration of red solution after dilution = c'

c=c'\times 5

1.4163\times 10^{-5} M=c'\times 5

c'=2.83\times 10^{-6} M

The final concentration of the diluted solution is 2.83\times 10^{-6} M

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