Answer:
0.36 M
Explanation:
The given are 15 grams of NaBr in a 400 mL solution. We need to calculate the Molarity.
Step 1: We know that molarity is:
[tex]Molarity = \frac{moles of solute}{Liter of solution}[/tex]
First, we need to get the equivalent moles of 15 grams of Sodium Bromide (NaBr) by using its molar mass. The molar mass of NaBr is:
Molar mass = 22.99 grams/mole Na + 79.90 grams / mole Br = 102.89 g/mole NaBr
moles of NaBr = 15 grams x [tex]\frac{1 mole}{102.89 grams} NaBr[/tex] = 0.1458 moles NaBr
Step 2: Convert the volume of solution from mL to Liter to avoid unit discrepancy. We know that 1000 mL = 1 Liter. Hence;
400 mL x [tex]\frac{1 Liter}{1000 mL}[/tex] = 0.4 Liter
Step 3: Substitute the derived values of moles of solute and volume of solution into the Molarity formula.
Molarity = [tex]\frac{0.1458 moles of NaBr}{0.4 L of solution}[/tex] = 0.36 M
Therefore, the Molarity of 15 grams NaBr dissolved in a 400 mL solution is 0.36M.
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