USA Biology Olympiad (USABO) Practice Exam

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If the molecular weight of fumarate is 160.0, how many grams are in 200 mL of a 0.1 M solution?

  1. 1.60 grams

  2. 3.20 grams

  3. 8.00 grams

  4. 16.00 grams

The correct answer is: 3.20 grams

To determine the mass of fumarate in a 200 mL solution at 0.1 M concentration, we can use the relationship between molarity (M), volume (V), and moles (n), which is given by the formula: \[ n = M \times V \] First, we need to convert the volume from milliliters to liters since molarity is expressed in moles per liter. 200 mL is equivalent to 0.2 liters. Next, we substitute the values into the formula: \[ n = 0.1 \, \text{M} \times 0.2 \, \text{L} = 0.02 \, \text{moles} \] Now, we need to calculate the mass in grams using the molecular weight of fumarate, which is 160.0 g/mol: \[ \text{mass} = n \times \text{molecular weight} \] \[ \text{mass} = 0.02 \, \text{moles} \times 160.0 \, \text{g/g/mol} = 3.2 \, \text{grams} \] The calculation clearly shows that the

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