Help with Chemistry! More than one Question?

Question

1. How many grams of sucrose, C12H22O11, must you use to make 150.00 mL of a 1.00 molar solution in Water?
2. What is the osmolarity of a 1.00 M solution of sucrose?
3. You have been given a 0.9% (m/v) NaCl solution. Answer the following:
a. What is the molarity of the above solution?
b. What is the osmolarity of the above solution?
c. Wht is the osmotic pressure of the NaCl solution in mmHg?
4. Given a 0.703 M NaCl solution that has a density of 1.0268 g/mL, calculate the following?
a. The % NaCl (m/m).
b. The molar concentration of NaCl.
5. Given the cryosopic constant of camphor is 37.8 degrees C/m. The symbole m is the molar concentration.
a. 0.1000 g of sulfur was dissolved in 10.000g of camphor. The freezing point of this solution dropped from 178.8 to 177.3 C. What is the molecular weight of sulfur?
b. Based on the answer above, how many sulfur atoms are in a sulfur molecule.

This is not a full homework assignment, this is the last 5 questions of a homework assignment. These are the ones I am having problems with.

I have looked over and read my text book numerous times, the problem is that I cannot find the information I am looking for. This covers many chapters and I am just trying to get someone to explain it. Even if I only get the equations that will be enough or an explination of how to get the answers that would be great too.

Answer 1

a)M. wt of sucrose = 342
1 mole of sucrose weighs 342 g
342 g of sucrose in 1000 ml is 1M soln
.342 g of sucrose in 1ml is 1M soln
(.342 x 150 ) = 51.3 g will give 150 ml of 1M soln
b)sucrose dissolves into water as sucrose. so its osmolarity is same as molarity.
3a)0.9 % mass/volume
or 0.9g NaCl in 100 ml of water
or 9g NaCl in 1000 ml of water
M.wt of NaCl = 58.5
molarity is 9/58.5 = 0.154 M
3b) NaCl dissociates into two parts (Na+ & Cl-)on dissolving. so it gives osmolarity of 2*(0.154)= 0.308
3c)for NaCl , vant hoff factor is 2.
M = 0.154M /L
R= 62.3637 L · mmHg · K-1 · mol-1
T= 298
osmotic pressure = i *M*R*T = 5723.98 mmHg
4a)1000 mL of solution weighs 1026.8 grams. it contains 0.703 moles of NaCl, which'll weigh 0.703 *58.5=41.125grams . so the rest of solution is water, i.e, 1026.8 - 41.125 =985.675 g is water
the moles of water is 985.675 / 18 (mol.wt.of water) = 54.76
so the mole/mole percent of NaCl is (0.703/54.76) * 100=1.283%
4b) i think its molal not molar as molar is already given. so here goes
1 litre weighs 1.0268kgand contains 0.703 moles NaCl
moles/kgof solution = 0.703/1.0268 = 0.685
5a) delta T = 178.8 - 177.3 =1.5 = i* 37.8*molality
molality is moles of sulfur/kg of solution
moles of sulfur = 0.1g/Mwt
molality is = [0.1g/Mwt]/[ 0.01kg camphor +0.001 kg of S] (im converting grams to kg)
we assume i =1, because camphor is a nonelectrolyte
so, equation becomes,
1.5 = i* 37.8*molality
= 37.8 *[0.1g/Mwt]/[ 0.01kg camphor +0.0001 kg of S]
= 37.8 *[0.1g/Mwt]/[ 0.0101kg ]
so M wt of S = 37.8 *[0.1g]/[ 0.0101kg *1.5] =249.5
5b) the atomic weight of atomic sulphur is 32.. so , number of atoms in a molecule in this case is 249.5/32 =7.79 or approx 8
so each of the molecules of sulfur in this case has 8atoms of sulfur

Answer 2

I think you really need to spend some time reading your textbook to find out how to do these problems. In number 5, you are wrong: m is the molal concentration, not the molar concentration. There is a very slight but important difference between these two that you will need to understand.

So, get out your book, read, look at the sample problems, and see if you can figure these out. If not, find a tutor or ask your instructor, your lab TA or a friend for help. You won't ever learn the material unless you work at it.

Answer 3

Is this an entire homework assignment?