Question

Determine the mass in grams of pentane that must be added to 67 g of benzene to make a 1.04 m solution. Give your answer to 2 decimal places.

Answer 1

To make a 1.04 m solution of pentane in 67 g of benzene, you would need 5.03 grams of pentane.

Step-by-step explanation:

To determine the mass in grams of pentane that must be added to 67 g of benzene to make a 1.04 m (molality) solution, we use the formula:

Molality (m) = moles of solute / kg of solvent

Since benzene is the solvent and we want a 1.04 m solution, we rearrange the formula to solve for the moles of pentane:

moles of pentane = molality × kg of benzene

First, convert the mass of benzene to kilograms:

67 g × (1 kg/1000 g) = 0.067 kg

Calculate the moles of pentane needed:

moles of pentane = 1.04 m × 0.067 kg = 0.06968 moles

Pentane (C5H12) has a molecular weight of about 72.15 g/mol. So, the mass of pentane needed is:

mass of pentane = moles of pentane × molecular weight of pentane

mass of pentane = 0.06968 moles × 72.15 g/mol = 5.027 grams

To two decimal places, the mass of pentane needed is 5.03 grams.

Answer 2

5.02 g pentane.

Step-by-step explanation:

Hello,

Molality is given by:

`m=(mol_(solute))/(mass_(solvent))`

Since the mass of the solvent (benzene) is 67 g, its value in kg is 0.067 kg because it is needed in that way in the formula. Now, solving for the moles of pentane (solute):

`mol_(solute)=m*m_(solvent)=1.04(mol)/(kg_(solvent))*0.067kg_(solvent)\\mol_(solute)=0.06968 mol`

Converting from moles to grams of pentane, we get the final answer:

`M_(pentane)=12*5+12=72g/mol\\\\m_(solute)=0.06968mol*(72g)/(1mol) =5.02gC_5H_(12)`

Best regards.