Question

an aqueous solution is 3.75% by mass of nonelectrolyte (MM=178g/mol). if the density of the solution is 1,012g/mL, calculate the molality of the solution, the mole fraction of the solute and the molarity of the solution.

Answer 1

Answer on question #80542, Chemistry-general chemistry

Question:

an aqueous solution is 3.75% by mass of nonelectrolyte (MM=178g/mol). if the density of the solution is 1,012g/mL, calculate the molality of the solution, the mole fraction of the solute and the molarity of the solution

p = 1,012g/cm^3

W=3,75%=0,037

MM = 178g/mol

Cm-?

mu -?

N-?

Solution:

We suggest V (solution = 1 L=1000ml. Theh, m(solution) =p*V=1,012*1000=1012g

m(soluted)= m(solution)*w=1012*0.037=37,4g

n=m(soluted)/Mm = 37,4/178=0.21 mol

Cm= n/V= 0.21/1 = 0.21mol/l

mu = n/1000g solvent

m(solvent)= m(solution)-m(soluted)= 1013-37.4 = 975,6 g

mu = 0.21*1000/975.6 = 0.215 mol/g

N = n/(n1+n2)

n(H2O) = m(solvent)/M(solvent) = 975,6/18 = 54,2mol

N = 0.21/(0.21+54,2) = 0.0038

Answer: Cm=0.21mol/l, mu = 0.215 mol/g, N=0.0038

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