Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
Who cares. Joe can solve his own problem instead of making people do it. (This was not toward you. This was supposed to be funny)
Answer:
I think it's Csl
I don't really know
but follow me and heart this
When the heat lost by water = heat gained by ethanol
∴( M* C * ΔT )w = (M*C*ΔT ) eth
when Mw mass of water = 40 g
C specific heat of water = 4.18
ΔT = (70- Tf)
and M(eth) mass (ethanol) = 40 g
C specific heat of ethanol = 2.44
ΔT = (Tf - 10 )
by substitution:
40* 4.18 * (70 - Tf) = 40 * 2.44 * (Tf-10)
∴Tf = 47.9 °C
Answer: Surface Tension
(Water molecules are attracted to the molecules in the wall of the glass beaker).
