Answer:
41.3 mL OF PENTANE CONTAINS 62.608 *10^-27 CARBON ATOMS.
Explanation:
Volume of Pentane = 41.3 mL
Density = 0.626 g/mL
Since density is defined as mass per unit volume. Calculate the mass:
Density = mass / volume
Mass = density * volume
Mass = 0.626 g/mL * 41.3 mL
Mass = 0.015 g of pentane
Molar mass of pentane C5H12 = ( 12* 5 + 1*12) = 60 + 12 = 72 g/mol
Number of moles = mass / molar mass
Number of moles = 0.015 g / 72 g/mol
Number of moles = 2.08 * 10^-4 moles.
Since 1 mole of pentane contains 5 moles of carbon, we can then calculate how many moles of csrbon 2.08*10^-4 moles will contain
1 mole of pentane = 5 moles of carbon
2.08 *10^-4 moles = x moles
x mole = 5 * 2.08*10^-4
x mole = 10.4 *10^-4 moles of carbon
In other words, 10.4*10^-4 moles of carbon is present in 41.3 mL of pentane
We can then calculate the number of carbon atoms present in the liquid using:
From 1 mole containing 6.02 *10^-23 atoms, 10.4 *10^-4 moles of pentane will contain:
1 mole = 6.02*10^-23 atoms
10.4 *10^-4 moles = (6.02*10^-23 * 10.4 *10^-4) atoms of pentane
= 62.608 *10 ^-27 atoms of pentane.
So therefore, 41.3 mL of pentane will contain 62.608 * 10 ^-27 carbon atoms
The number of C atoms in 41.3 mL of pentane (C₅H₁₂) of density 0.626 g/mL is 2.16x10²³.
To find the number of C atoms in pentane, we need to find the number of moles.
We can calculate the mass from the density:
[tex] d = \frac{m}{V} [/tex]
Where:
d: is the density = 0.626 g/mL
V: is the volume = 41.3 mL
The mass is:
[tex] m = d*V = 0.626 g/mL*41.3 mL = 25.85 g [/tex]
Now, with the molecular weight of the pentane, we can calculate the number of moles:
[tex] \eta = \frac{m}{M} = \frac{25.85 g}{72.15 g/mol} = 0.358 \:moles [/tex]
The number of C atoms can be found with Avogadro's number:
[tex] n_{C} = \frac{6.022\cdot 10^{23} \:atoms}{1\:mol}*0.358 \:moles = 2.16 \cdot 10^{23} \:atoms [/tex]
Therefore, the liquid contains 2.16x10²³ C atoms.
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