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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.

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