Question

How many grams of sodium hydroxide are present in 247.0 mL of a 0.300 M NaOH solution?

Answer 1

There are 2.964 grams of sodium hydroxide in 247.0 mL of a 0.300 M NaOH solution.

Step-by-step explanation:

To calculate how many grams of sodium hydroxide are present in 247.0 mL of a 0.300 M NaOH solution, we need to use the concept of molarity (M), which is moles of solute per liter of solution.

First, we convert the volume from milliliters to liters:

247.0 mL × (1 L / 1000 mL) = 0.247 L

Next, we use the molarity to find the moles of NaOH:

0.300 M = 0.300 moles/L × 0.247 L = 0.0741 moles of NaOH

Lastly, we convert moles to grams using the molar mass of NaOH (approximately 40 g/mol):

0.0741 moles × 40 g/mol = 2.964 g

Therefore, there are 2.964 grams of sodium hydroxide in the given solution.

Answer 2

The grams of NaOH that are present in 247.0 ml of a 0.300M NaOH solution is 2.964 grams

calculation

Step 1: find moles of NaOH

moles = molarity x volume in liters

volume in liters = 247.0 /1000 = 0.247 liters

molarity = 0.300 M = 0.300 mol/l

moles = 0.300 mol/l x 0.247 =0.0741 moles

step 2: find mass of NaOH

mass= moles x molar mass

The molar mass of NaOH = 23 +1 +16= 40 g/mol

mass=0.0741 moles x40 g/mol =2.964 grams