Freezing Point - Physical Chemistry

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Question

When 20 grams of a solute is added to a liter of water, the freezing point becomes What is the molar mass of the solute?

Assume that the solute does not make ions in solution.

$K_$f=1.86$\frac{^{\circ}$$C}{m}$

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Answer

We can use the freezing point depression equation in order to determine the molar mass of the unknown solute.

$\Delta $T_{f}$ = $K_{f}$mi$

$2.3 = $(1.86$\frac{^{\circ}$$C}{m})($\frac{\frac{20grams}{x}$}{1 liter})(1)$

$x = 16.2$\frac{g}{mol}$$

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Question

When 20 grams of a solute is added to a liter of water, the freezing point becomes What is the molar mass of the solute?

Assume that the solute does not make ions in solution.

$K_$f=1.86$\frac{^{\circ}$$C}{m}$

Tap to reveal answer

Answer

We can use the freezing point depression equation in order to determine the molar mass of the unknown solute.

$\Delta $T_{f}$ = $K_{f}$mi$

$2.3 = $(1.86$\frac{^{\circ}$$C}{m})($\frac{\frac{20grams}{x}$}{1 liter})(1)$

$x = 16.2$\frac{g}{mol}$$

← Didn't Know|Knew It →

Question

When 20 grams of a solute is added to a liter of water, the freezing point becomes What is the molar mass of the solute?

Assume that the solute does not make ions in solution.

$K_$f=1.86$\frac{^{\circ}$$C}{m}$

Tap to reveal answer

Answer

We can use the freezing point depression equation in order to determine the molar mass of the unknown solute.

$\Delta $T_{f}$ = $K_{f}$mi$

$2.3 = $(1.86$\frac{^{\circ}$$C}{m})($\frac{\frac{20grams}{x}$}{1 liter})(1)$

$x = 16.2$\frac{g}{mol}$$

← Didn't Know|Knew It →

Question

When 20 grams of a solute is added to a liter of water, the freezing point becomes What is the molar mass of the solute?

Assume that the solute does not make ions in solution.

$K_$f=1.86$\frac{^{\circ}$$C}{m}$

Tap to reveal answer

Answer

We can use the freezing point depression equation in order to determine the molar mass of the unknown solute.

$\Delta $T_{f}$ = $K_{f}$mi$

$2.3 = $(1.86$\frac{^{\circ}$$C}{m})($\frac{\frac{20grams}{x}$}{1 liter})(1)$

$x = 16.2$\frac{g}{mol}$$

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