A 2.400×10 2 M solution of NaCl in water is at 20.0 C . The sample was created b

Question

A 2.400×102 M solution of NaCl in water is at 20.0 C . The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L . It was determined that the volume of water needed to do this was 999.4 mL . The density of water at 20.0 C is 0.9982 g/mL .

1)Calculate the molality of the salt solution.

2)Calculate the mole fraction of salt in this solution.

3)Calculate the concentration of the salt solution in percent by mass.

4)Calculate the concentration of the salt solution in parts per million.

Explanation / Answer

Molarity = 2.400×102 M

volume = 1 L

moles of NaCl = M x V = 2.400×102

weight of water = d x V

                         = 0.9982 x 999.4

                          = 997.6 g

                          = 0.9976 kg

(1) molality = moles of solute / solvent weight in kg

                     = 2.400×102 /0.9976

                     = 0.2405m

(2)

X1 = mole fraction of NaCl , X2 = mole fraction of H2O

m = molality

m2 = molar mass of NaCl

then formula

X1 / X2 = m x m2 / 1000

X1/ 1-X1 = m x m2 /1000

X1 / 1-X1 = 0.2405 x 58.5 / 1000

X1 = 0.014

mole fraction of NaCl = 0.014

(3) moles of NaCl = 2.400×102

mass of NaCl = moles x molar mass = 2.400×102 x 58.5 = 1.404 g

water mass = 997.6 g

total solution mass = 997.6 g + 1.404 = 999.004 g

salt mass % = (1.404 / 999.004 ) x 100

                      = 0.14 %

(4) concentration in ppm = (1.404 / 999.004 ) x 10^6

                                        = 1405 ppm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.