A beaker contains 100 grams of 1.20 M NaCl. If you transfer 50% of the solution to another beaker, what is the molarity of the solution remaining in the first beaker?
Think about it a little. Suppose we have a beaker containing a 1 M solution of NaCl. Suppose the beaker contains 1 L of solution; therefore, we know it contains 1 mol/L or 58.5(approximately) g NaCl/1000 mL.
Now let's pour 500 mL out onto the floor. What do we have left in the beaker? We have 0.50 mole NaCl (since we lost half of it) and we have 500 mL left (0.5L since we lost 0.5L) so the concn is not 0.5mol/0.5L and voila, the solution is still 1.0 M. Let's take the same beaker we had at the beginning and pour 900 mL on the floor. That leaves us 100 mL in the beaker(the other 900 is on the floor) and it leaves 0.1 mole NaCl(the other 0.9 is on the floor). So the concn of th solution remaining in the beaker is 0.1mole/0.1L = voila, 1.0M. Think about it again. Do you think simply pouring some of the solution out of one container into another container will change the concn. It's true you lose some of the solute (the NaCl in this case) but it's also true that you lose the same amount of solvent, so the concn remains exactly the same.
To determine the molarity of the solution remaining in the first beaker, you need to consider that molarity is defined as the amount of solute (in moles) divided by the volume of the solution (in liters).
In this case, the initial beaker contains 100 grams of a 1.20 M NaCl solution. To find the number of moles of NaCl, you can use the molar mass of NaCl, which is 58.44 g/mol. Dividing the mass (100 grams) by the molar mass gives you the number of moles:
100 g / 58.44 g/mol = 1.71 mol
Next, consider that you are transferring 50% of the solution to another beaker. This means that you are transferring half of the volume of the solution. Since molarity is directly proportional to the volume of the solution, the molarity of the remaining solution will also be reduced by 50%.
Therefore, the molarity of the solution left in the first beaker would be:
1.20 M * 0.50 = 0.60 M
So, the molarity of the solution remaining in the first beaker is 0.60 M.
To find the molarity of the solution remaining in the first beaker after transferring 50% of it to another beaker, you can follow these steps:
Step 1: Calculate the amount of NaCl transferred to the second beaker:
The original beaker contains 100 grams of NaCl, and if we transfer 50% of it, we have:
Amount transferred = 100 grams × 0.5 = 50 grams of NaCl.
Step 2: Determine the amount of NaCl remaining in the first beaker:
The original beaker initially had 100 grams of NaCl, and we transferred 50 grams to the second beaker. So, the remaining amount is:
Amount remaining = 100 grams - 50 grams = 50 grams of NaCl.
Step 3: Calculate the volume of the solution remaining in the first beaker:
Since we transferred only the solute (NaCl) to the second beaker, the volume of the solution in the first beaker remains the same. Hence, there is no change in volume.
Step 4: Find the molarity of the solution remaining in the first beaker:
Molarity (M) is defined as the moles of solute per liter of solution. To calculate the molarity, we first need to convert the grams of solute to moles of solute using the molar mass of NaCl, which is 58.44 g/mol:
moles of NaCl remaining = 50 grams / 58.44 g/mol = 0.856 mol.
Since there is no change in volume, the amount of solution (in liters) in the first beaker is still the same.
Therefore, the molarity of the solution remaining in the first beaker is 0.856 M.