Question

how many moles of iodine should be added to 750 grams of carbon tetrachloride to prepare a 0.24 m solution

Answer 1

Answer:

0.18 mol

Explanation:

Given data

• Mass of carbon tetrachloride (solvent): 750 g

• Molality of the solution: 0.24 m

• Moles of iodine (solute): ?

Step 1: Convert the mass of the solvent to kilograms

We will use the relationship 1 kg = 1,000 g.

$750g \times \frac{1kg}{1,000g} =0.750kg$

Step 2: Calculate the moles of the solute

The molality is equal to the moles of solute divided by the kilograms of solvent. Then,

$m = \frac{moles\ of\ solute }{kilograms\ of\ solvent} \\moles\ of\ solute = m \times kilograms\ of\ solvent = \frac{0.24mol}{kg} \times 0.750kg = 0.18 mol$

Answer 2

Answer:

$n=0.18mol$

Explanation:

Hello,

In this case, we considering the iodine as the solute and the carbon tetrachloride as the solvent, then for the given molal solution which is measured in moles of solute per kilograms of solvent we first compute the required kilograms:

$m_{solvent}=750g*\frac{1kg}{1000g} =0.750kg$

Then, we can compute the moles:

$n=0.750kg*0.24\frac{mol}{kg}\\ \\n=0.18mol$

Best regards.