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2 years ago

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Calculate the molality of a solution that contains 51.2 g of naphthalene, C10H8, in 500 mL of carbon tetrachloride. The density

of CCl4 is 1.60 g/mL.

1 answer:

PtichkaEL [24]2 years ago

4 0

<u>Answer:</u> The molality of naphthalene solution is 0.499 m

<u>Explanation:</u>

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Volume of carbon tetrachloride = 500 mL

Density of carbon tetrachloride = 1.60 g/mL

Putting values in equation 1, we get:

$\text{Mass of carbon tetrachloride}=(1.60g/mL\times 500mL)=800g$

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molarity:

$\text{Molality of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (in g)}}$ .....(2)

Given values:

Given mass of naphthalene = 51.2 g

Molar mass of naphthalene = 128.17 g/mol

Mass of solvent = 800 g

Putting values in equation 2, we get:

$\text{Molality of naphthalene}=\frac{51.2\times 1000}{128.17\times 800}\\\\\text{Molality of naphthalene}=0.499m$

Hence, the molality of naphthalene solution is 0.499 m

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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

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Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

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$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

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The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

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$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

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Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

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