13 of 24 A 2.900×10-2 M solution of NaCl in water is at 20.0° C. The sample was

Question

13 of 24 A 2.900×10-2 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 999.2 mI. The density of water at 20.0° C is 0.9982 g/mL. Part A Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units. View Available Hint(s) mNac- 1700 Units Submit Part B Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures. View Available Hint(s)

Explanation / Answer

A)
volume , V = 1 L


use:
number of mol ,
n = Molarity * Volume
= 0.02900*1
= 2.900*10^-2 mol
volume , V = 1 L
= 1*10^3 mL


density, d = 0.9982 g/mL
use:
mass = density * volume
= 0.9982 g/mL *1*10^3 mL
= 998.2 g
This is mass of solution

Molar mass of NaCl,
MM = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol

use:
mass of NaCl,
m = number of mol * molar mass
= 2.900*10^-2 mol * 58.44 g/mol
= 1.695 g
This is mass of solute

mass of solvent = mass of solution - mass of solute
= 998.2 - 1.695
= 996.5 g
= 0.9965 Kg

m(solvent)= 0.9965 Kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(2.9*10^-2 mol)/(0.9965 Kg)
= 2.910*10^-2 molal
Answer: 2.910*10^-2 molal

B)
mol of NaCl = 2.900*10^-2 mol

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass(H2O)= 996.5 g

use:
number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(996.5 g)/(18.016 g/mol)
= 55.31 mol

Mole fraction = mol of NaCl / (mol of NaCl + mol of H2O)
= (2.900*10^-2) / (2.900*10^-2 + 55.31)
= 5.240*10^-4
Answer: 5.240*10^-4

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