Answer:
1.714 M
Explanation:
We'll begin by calculating the number of mole in 46.8 g of NaHCO₃. This can be obtained as follow:
Mass of NaHCO₃ = 46.8 g
Molar mass of NaHCO₃ = 23 + 1 + 12 + (3×16)
= 23 + 1 + 12 + 48
= 84 g/mol
Mole of NaHCO₃ =?
Mole = mass / molar mass
Mole of NaHCO₃ = 46.8 / 84
Mole of NaHCO₃ = 0.557 mole
Next, we shall convert 325 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
325 mL = 325 mL × 1 L / 1000 mL
325 mL = 0.325 L
Thus, 325 mL is equivalent to 0.325 L.
Finally, we shall determine the molarity of the solution. This can be obtained as shown below:
Mole of NaHCO₃ = 0.557 mole
Volume = 0.325 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.557 / 0.325
Molarity = 1.714 M
Therefore the molarity of the solution is 1.714 M
