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3 years ago

How many atoms of gold are in gold nugget with a mass of 4.25 g

Chemistry

2 answers:

Norma-Jean [14]3 years ago

7 0

Answer:

1.299 * 10^22

Explanation:

Feliz [49]3 years ago

3 0

Number of Atoms in Gold for given mass can be calculated using following formula,

# of Moles = Number of Atoms / 6.022 × 10²³
Or,
Number of Atoms = Moles × 6.022 × 10²³ ------- (1)

Calculating Moles,
As,
Moles = Mass / M.mass
So,
Moles = 4.25 g / 196.96 g/mol

Moles = 0.0215

Putting value of mole in eq.1,

Number of Atoms = 0.0215 × 6.022 × 10²³

Number of Atoms = 1.299 × 10²²

Result:
4.25 g of Gold Nugget contains 1.299 × 10²² Atoms.

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Alexxx [7]

Answer: The vapor pressure of solution is 459.17 mmHg

Explanation:

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$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

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Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol$

For diethyl ether:

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol$

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$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}$

$\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095$

The formula for relative lowering of vapor pressure will be:

$\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}$

where,

$p^o$ = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

$p^s$ = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

$\chi_{\text{solute}}$ = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

$\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg$

Hence, the vapor pressure of solution is 459.17 mmHg

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