A beaker contains 41.3 mL of pentane ( C5H12 , density is 0.626 g/mL). Determine how many C atoms this liquid contains.

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