If you have 1.00 M sodium chloride solution, and you have 300. mL of it, how many grams of sodium chloride are there?
molarity=(:mass/molmass)/volume
mass=molarity*molmass*volume=58.4*1.0*.300=17.5grams
To find the number of grams of sodium chloride in a given solution, you need to know the concentration of the solution in terms of molarity (moles of solute per liter of solution) and the volume of the solution.
Here are the steps to determine the number of grams of sodium chloride in a 1.00 M sodium chloride solution:
1. Determine the molarity of the solution. In this case, the given solution is a 1.00 M sodium chloride solution, which means there is 1.00 mole of sodium chloride present in one liter of the solution.
2. Since the volume of the solution is given in milliliters (mL), you need to convert it to liters. There are 1000 mL in 1 liter, so 300 mL is equal to 0.300 L.
3. Use the molarity and volume to calculate the amount of moles of sodium chloride in the solution. Multiply the molarity by the volume in liters:
Moles of NaCl = Molarity x Volume
= 1.00 M x 0.300 L
= 0.300 moles of NaCl
4. Finally, calculate the grams of sodium chloride by multiplying the moles of sodium chloride by the molar mass of sodium chloride, which is 58.44 grams per mole (g/mol):
Grams of NaCl = Moles of NaCl x Molar mass
= 0.300 moles x 58.44 g/mol
= 17.53 grams of NaCl
Therefore, there are approximately 17.53 grams of sodium chloride in 300 mL of a 1.00 M sodium chloride solution.