what is the boiling point of a 0.321 m aqueous solution of NaCl. Enter your rounded answer with 3 decimal places.

∆T = imK

∆T = Change in boiling point (B.P.)

i = van't Hoff factor = 2 for NaCl (Na+ and Cl2_

m = molality = 0.321 m

k = boiling point constant for water = 0.512 deg/m

∆T = (2)(0.321)(0.512) = 0.329 degrees

Since the normal B.P. for water is 100ºC, the new boiling point of this solution is 100 + 0.329 = 100.329ºC

So Our Answer is 100.329ºC

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.