4 Added MathJax syntax edited Aug 5 '17 at 17:05 andselisk♦ 22.2k88 gold badges7878 silver badges149149 bronze badges A 4.305-g sample of a nonelectrolyte is dissolved in 105 g of water. The solution freezes at -1.23C. Calculate the molar mass of the solute. Kf for water = 1.86C/m. A $$\pu{4.305 g}$$ sample of a nonelectrolyte is dissolved in $$\pu{105 g}$$ of water. The solution freezes at $$\pu{-1.23^\circ C}$$. Calculate the molar mass of the solute. $$k_f$$ for water is $$\pu{1.86 C/m}$$. My steps: 1.23=1.86 * x/1.05 where x=y/4.305 0.66=x/1.05 x=0.69$$1.23 = 1.86 \times \frac{x}{1.05},$$ 0.69=y/4where $$x = y / 4.305$$.305 y=2.989 g$$0.66 = \frac{x}{1.05}$$ $$x = 0.69$$ $$0.69 = y / 4.305$$ $$y = \pu{2.989 g}$$ But I know this is not correct. A 4.305-g sample of a nonelectrolyte is dissolved in 105 g of water. The solution freezes at -1.23C. Calculate the molar mass of the solute. Kf for water = 1.86C/m. My steps: 1.23=1.86 * x/1.05 where x=y/4.305 0.66=x/1.05 x=0.69 0.69=y/4.305 y=2.989 g But I know this is not correct. A $$\pu{4.305 g}$$ sample of a nonelectrolyte is dissolved in $$\pu{105 g}$$ of water. The solution freezes at $$\pu{-1.23^\circ C}$$. Calculate the molar mass of the solute. $$k_f$$ for water is $$\pu{1.86 C/m}$$. My steps: $$1.23 = 1.86 \times \frac{x}{1.05},$$ where $$x = y / 4.305$$. $$0.66 = \frac{x}{1.05}$$ $$x = 0.69$$ $$0.69 = y / 4.305$$ $$y = \pu{2.989 g}$$ But I know this is not correct. 3 edited tags | link edited Jul 19 '14 at 2:43 ron 73.2k1010 gold badges161161 silver badges267267 bronze badges 2 edited tags | link edited Jul 18 '14 at 22:49 James 611 gold badge11 silver badge22 bronze badges 1 asked Jul 18 '14 at 22:44 James 611 gold badge11 silver badge22 bronze badges