In a laboratory experiment, a 10.0-mL sample of NaCl solution is poured into an evaporating dish with a mass of 24.10 g. The combined mass of the evaporating dish and NaCl solution is 36.15 g. After heating, the evaporating dish and dry NaCl have a combined mass of 25.50 g.
b. What is the molarity (M) of the NaCl solution?

Question

In a laboratory experiment, a 10.0-mL sample of NaCl solution is poured into an evaporating dish with a mass of 24.10 g. The combined mass of the evaporating dish and NaCl solution is 36.15 g. After heating, the evaporating dish and dry NaCl have a combined mass of 25.50 g.
b. What is the molarity (M) of the NaCl solution?

b. What is the molarity (M) of the NaCl solution?

Accepted Answer

Welcome back, everyone in a chemical laboratory. A task was assigned to analyze a molar of picric acid C six H three N eight N 307 solution. A Hager's re agent for detecting alkaloids in an empty evaporating dish weighing 29.5 g. A 20 mL sample of pic acid solution is placed the total mass of the evaporating dish containing the p acid solution is 49.7 g. The picric acid solution placed in the evaporating dish is dried in an oven and after drawing a total mass of 29.7 g of tamed for the evaporating dish. And the anhydrous picric acid determine the molar of the picric acid solution. We're going to need to begin by recalling that molar of our solution can be determined by taking our moles of our solute. In this case, that would be our dried picric acid. So our anhydrous picric acid C six H three N 307, which is then divided by our volume of our solution in liters. Now, in order to find our numerator which is for moles of our solute and hydrous picric acid, we're going to need to find our mass of anhydrous picker acid C six H three N 307. And in order to determine the mass of our dried picric acid, we're going to need to take the difference between the mass of our evaporating dish which contains the anhydrous picric acid plus the anhydrous picric acid within the dish. And then we want to subtract from the mass of our evaporating dish again, which contains this anhydrous picric acid. So, plugging in these two terms, our mass of our evaporating dish and anhydrous picric acid was told to be 29.7 g. So we'll plug that in as 29. g. And then this is subtracted from our mass of our evaporating dish which from the prompt is also given to us for its empty condition as 29.5 g. So taking that mass of the empty evaporating dish, we have 29.7 g minus 29.5 g. And from this difference, we'll find our mass of our N hydros picric acid equal to a value of 0.2 g. Now, we're going to be able to go into moles of our and hydris picric acid in which we're going to begin with that mass. We just found 0.2 g of picric acid N 307 or C six H three N 307. So just so everything is visible and then we're going to multiply it by a conversion factor to cancel out grams in our denominator and get to moles in our numerator and using our periodic table and the chemical formula of picric acid. We're going to determine the molecular weight of picric acid in which we would find that for a mass of 229. g of picric acid, we have this amount of mass per mole of picric acid. So now canceling out grams, we're left with moles as our final unit. And we would find that our moles of our anhydrous picric acid is equal to 8.730 times to the negative fourth power moles of our anhydrous picric acid. So now with moles of our anhydrous solute and from the prompt, our given volume of our solution, which was given as 20 mL. We're going to be able to find the molar. Now recall that in our molar solution volume is in or in our molar formula volume is in liters. So we're going to need to take that given volume from the prompt for 20 mL of solution. And we're going to multiply to go from milliliters in the denominator to liters in the numerator. Recall that our prefix Milly tells us we have an equivalent of 10th of the negative third of our base unit liter. So canceling out millis we're left with liters as our unit of volume and this will simplify so that we find our volume of our solution equal to 0.2 liters of solution. And now going into calculating our molar of our anhydrous solution correction of our actual solution of picric acid, which does contain water. We would take our moles of our anhydrous solute, which we determined again to be 8.730 times 10 to the negative fourth power moles of anhydrous picric acid. And this is divided by our volume of our solution, which we just determined above as 0.2 liters of solution. So that leaves us with moles per liter as our final unit. And in our next line, we'll find that our molar of our solution is equal to 0.436 molar. And this would be our final answer as our polarity of our solution of picric acid. So I hope that this made sense and let us know if you have any questions.

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Key Concepts

Molarity

Mass and Moles Relationship

Evaporation and Residue