You are watching: Compute the freezing point of this solution 25.5 g c7h11no7s
$$1.23 = 1.86 imes fracx1.05,$$
where $x = y / 4.305$.
$$0.66 = fracx1.05$$$$x = 0.69$$$$0.69 = y / 4.305$$$$y = pu2.989 g$$
But I recognize this is not correct.
The freezing point depression is based upon the molal concentration (moles that solute every kg that solvent).
$$Delta T_f = -k_f cdot m$$
You recognize the freezing suggest depression the the solution and also the cryoscopic constant, for this reason you have the right to calculate the molality:
$$m = -dfracDelta T_fk_f=-dfracpu-1.23 ^circ C1.86 frac extrmkgcdot ^circ extC extrmmol=pu0.66 molal$$
Since molality is the number of moles the solute divided by the mass of the solvent in kg, you can calculate the moles of the solute:
$$n_ extrmsolute= pu0.66 m cdot pu 0.105 kg= ...$$
Divide your mass of solute through the mole of solute and you will get the molar massive (and no 2.989 g).
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edited might 25 in ~ 6:40
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answered Jul 19 "14 at 1:19
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