Question

1 pts

What is the molarity when 8.4 grams of NaF (molar mass = 42 grams/mol) is placed in 100
ml of water?

Answer 1

The molarity of a NaF solution with 8.4 grams of NaF dissolved in 100 ml of water is calculated by dividing the number of moles of NaF (0.2 mol) by the volume of the solution in liters (0.1 L), which results in a molarity of 2.0 M.

Step-by-step explanation:

To calculate the molarity of a NaF solution when 8.4 grams of NaF (molar mass = 42 grams/mol) is dissolved in 100 ml of water, follow these steps:

1. Calculate the number of moles of NaF by dividing the mass of NaF by its molar mass:

\(\frac{8.4 \text{ g}}{42 \text{ g/mol}} = 0.2 \text{ mol}\)

1. Convert the volume of the solution from milliliters to liters:

100 ml = 0.1 L

1. Use the definition of molarity (M), which is moles of solute per liter of solution:

\(M = \frac{0.2 \text{ mol}}{0.1 \text{ L}}\)

1. Complete the calculation to find the molarity of the NaF solution:

M = 2.0 M

Answer 2

molarity = 2 mol/ L

Step-by-step explanation:

We know that molarity is equal to the number of moles of solute divided by volume of solution in litter.

Given data:

mass of NaF = 8.4 g

molar mass of NaF = 42 g/mol

volume of water = 100 mL

First of all we will convert the milliliter of solution in to litter.

we know that 1 litter is equal to 1000 mL.

100/ 1000= 0.1 L

Formula:

molarity = moles/ volume in litter

Now we will calculate the number of moles.

number of moles = mass/ molar mass

number of moles = 8.4 g/ 42 g/mol

number of moles = 0.2 mol

By putting the values in molarity formula, we will get,

molarity = 0.2 mol/ 0.1 L

molarity = 2 mol/ L