Asolution of salt (molar mass 90 g mol-1) in water has a density of 1.29 g/ml. the concentration of the salt is 35% by mass.

a. calculate the molality of the solution.

b. calculate the molarity of the solution.

c. calculate the mole fraction of the salt in the solution.

a) 5.9846 mol/kg

b) 5.018 mol/L

c)0.09725

Explanation:

consider 100 g of solution

now

since the salt is 35 % , water will be 65 %

now

mass of salt = 35 g

mass of water = 65 g

we know that

moles = mass / molar mass

so

moles of salt = 35 g / 90 (g/mol) = 0.389 mol

moles of water = 65 / 18 = 3.6111 mol

now

volume of solution = mass of solution / density of solution

volume of solution = 100 g / ( 1.29 g/ml)

volume of solution = 77.52 ml

volume of solution = 0.07752 L

a)molality = moles of salt / mass of water (kg)

molality = 0.389 mol / 0.065 kg

molality = 5.9846 mol/kg

b)

molarity = moles of salt / volume of solution (L)

molarity = 0.389 mol / 0.07752 L

molarity = 5.018 mol/L

c)

now

total moles in the solution = moles of salt + moles of water

total moles in the solution = 0.389 + 3.6111

total moles in the solution = 4 mol

now

mole fraction of salt = moles of salt / total moles

mole fraction of salt = 0.389 mol / 4 mol

mole fraction of salt = 0.09725

the answer is c i believe