A $pu4.305 g$ sample that a nonelectrolyte is liquified in $pu105 g$ that water. The systems freezes at $pu-1.23^circ C$. Calculate the molar mass of the solute. $k_f$ because that water is $pu1.86 C/m$.

You are watching: Compute the freezing point of this solution 25.5 g c7h11no7s

My steps:

$$1.23 = 1.86 imes fracx1.05,$$

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.


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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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Buck Thorn♦
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answered Jul 19 "14 at 1:19
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