*To do this, we need the molecular weight of glucose.*

*To do this, we need the molecular weight of glucose.*

Next, we need to convert the denominator from milliliters to liters, which we do through the unit definition that there are 1000 milliliters per liter of solution.

Carrying out this math, we get a value of 0.111, and canceling units, we see that the final units are moles of glucose per liter of solution. Note that since the experimental conditions were known to three significant figures (both the 2.00 grams and the 100 milliliters are given to three significant figures), we keep three significant figures in the final result.

(Glucose is a simple sugar, with a molecular formula C.) Our goal is to calculate the concentration in moles per liter or, equivalently, to determine the molarity of the solution. We have 2.00 grams of glucose per 100 m L of solution.

First, we convert from grams of glucose to moles of glucose.

From the findings, the difficulties identified were lack of understanding of the mole concept, inability to balance chemical equations, use of inconsistent stoichiometric relationships, identifying the limiting reagent, determination of theoretical yields and identification of substances in excess.

The study also found that the use of problem-solving instruction as effective in remedying the identified difficulties in comparison to the conventional lecture method.

It is defined as the moles of a substance contained in one liter of solution.

For instance, if a solution has a concentration of 1.20 M Na Cl, this means that there are 1.20 moles of Na Cl per liter of solution.

Step #2 Using the equation, compare the moles of water made from moles of ethane.

Step #3 Convert the moles of ethane back into grams of ethane.

## Comments Solving Stoichiometry Problems