Questions: How many grams of sodium Fluoride are in 700 mL of 3.0 M of solution of sodium fluoride, NaF Molar mass = 42 g / mol? (A) 88 g (B) 67 g (C) 9.8 g (D) 2.1 g (E) 26 g

Solution Steps

To find the number of moles of sodium fluoride (NaF) in the solution, we use the molarity formula: \[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] Given:

• Molarity (M) = 3.0 M
• Volume of solution = 700 mL = 0.700 L

Rearranging the formula to solve for moles of solute: \[ \text{moles of NaF} = \text{Molarity} \times \text{Volume} \] \[ \text{moles of NaF} = 3.0 \, \text{M} \times 0.700 \, \text{L} = 2.1 \, \text{moles} \]

To find the mass of NaF, we use the molar mass of NaF: \[ \text{Mass} = \text{moles} \times \text{molar mass} \] Given:

• Molar mass of NaF = 42 g/mol

\[ \text{Mass of NaF} = 2.1 \, \text{moles} \times 42 \, \text{g/mol} = 88.2 \, \text{g} \]

Since the options provided are in whole numbers, we round 88.2 g to the nearest whole number: \[ 88.2 \, \text{g} \approx 88 \, \text{g} \]

Final Answer