Chemist. help?

Question

1) the molality of an aqueous solution of sodium chloride( NaCl) is 3. 6 m and the density of the same is 1.1g/mL.
a) What is the solute and Solvent in this solution?
b) A 3.6 molality aqueous solution of NaCl contains ______moles of NaCl in ______ of _____.
c) how many grams of NaCl are contained in 3.6moles of NaCl?
d) Take 3.6moles of NaCl and dissolve them in 1000g of water. what is the total weight of this solution?
e) calculate the weight percentage of NaCl in this solution.
f) How many moles of water do you have in 1000g of this solvet?
g) Calculate the molar fraction of sodium chloride in a 3.6 molality aqueous solution of the salt.
h) the density of the solution is 1.2g/mL, what is the total volume of the solution . total weight of solution =
i) solve the molar concentration of NaCl in the solution that you prepared above.
j) ^T=iKcm, calculate the change in the freezing point and the boiling point of a 3.6 molality aqueous solution of NaCl (Kb:0.51degre,Kf1.86

Answer 1

a) NaCl is the solute, this is what you are dissolving
H2O is the solvent, this is what is doing the dissolving

b) If the solution is 3.6 molal, then by the definition of molality, this solution has 3.6 moles solute (NaCl)/ 1 kilogram solvent (H2O)

c) 3.6 moles NaCl x 58.44 g
..........................--------------
............................1 mol
210.38 g NaCl

d) If 3.6 moles weighs 210.38 g and you have 1000 g of water, add the two together, and you get 1210.38 g total weight

e) Mass percent is mass of the part/mass total x 100
210.38g/1210.38g x 100 = 17.38%

f) 1000g H2O converted to moles with the same procedure as converting NaCl to moles gives you 55.49 moles H2O

g) Mole fraction = moles NaCl
.......................---------------------
...................... .total moles

= 3.6 moles NaCl
-------------------------
3.6 + 55.49

=.0609

h) Density=mass/volume so
volume = mass/density
=1210.38 g/1.2 g/mL
=1100.34 mL solution

i) M=moles solute/L solution
=3.6 moles/1.10034 L
=3.27 M

j) For the freezing point depression
delta T=Kf x m x i
=1.86 x 3.6 m x 2
=13.39 degrees C

For the boiling point elevation
delta T=Kb x m x i
=.51 x 3.6 m x 2
=3.67 degrees C

Hope this helps! Sorry if I made any mistakes