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3. To find the grams of NaCl needed, we need to use the formula:

molarity (M) = moles of solute / liters of solution

First, we need to calculate the moles of NaCl:

3.46 mol/L = moles of NaCl / 1 L

moles of NaCl = 3.46 mol/L

Next, we need to convert the moles of NaCl to grams:

moles of NaCl = (mass of NaCl in grams) / (molar mass of NaCl)

mass of NaCl in grams = (moles of NaCl) x (molar mass of NaCl)

The molar mass of NaCl is 58.44 g/mol.

mass of NaCl in grams = (3.46 mol/L) x (58.44 g/mol) = 202.21 g/L

To make a 3.46 M solution, we would need 202.21 g of NaCl per liter of solution. Since we want to make a 100. g solution, we will need to use a proportion:

202.21 g / L = x g / 0.1 L

x = 20.22 g

Therefore, we need 20.22 g of NaCl to dissolve in 100. g of water to make a 3.46 m solution.

4.

(a) Mass percent of CH3OH:

mass percent = (mass of solute / mass of solution) x 100%

mass of solute = 16.0 g CH3OH

mass of solution = 16.0 g CH3OH + 50.0 g water = 66.0 g

mass percent of CH3OH = (16.0 g / 66.0 g) x 100% = 24.2%

(b) Mole fraction of CH3OH:

Mole fraction = (moles of component / total moles in solution)

First, we need to calculate the moles of CH3OH:

moles of CH3OH = 16.0 g / (32.04 g/mol) = 0.499 mol

Next, we need to calculate the moles of water:

moles of H2O = 50.0 g / (18.02 g/mol) = 2.777 mol

Total moles in solution = moles of CH3OH + moles of H2O = 3.276 mol

Mole fraction of CH3OH = 0.499 mol / 3.276 mol = 0.152

Mole fraction of H2O = 2.777 mol / 3.276 mol = 0.848

(c) Molarity and molality of CH3OH:

Molarity (M) = moles of solute / liters of solution

moles of CH3OH = 0.499 mol

liters of solution = (16.0 g + 50.0 g) / (0.955 g/mL) = 66.49 mL = 0.06649 L

Molarity of CH3OH = 0.499 mol / 0.06649 L = 7.50 M

Molality (m) = moles of solute / kilograms of solvent

moles of CH3OH = 0.499 mol

kilograms of solvent = 50.0 g / 1000 g/kg = 0.0500 kg

Molality of CH3OH = 0.499 mol / 0.0500 kg = 9.98 m