The mass of a dry, 50 mL beaker is 49.135 g. The observed volume of water in a volumetric pipet is 20.00 mL. This water is transferred to the beaker and the combined mass of the water and the beaker is found to be 69.122 g. Given that the density of water at 24 degrees Celsius is 0.99732 g/mL, calculate:

a) The mass of water:

b) The actual volume of water in the beaker:

c)Percent error in the volume measurement:

Really need some help solving these please.

You are too impatient. 15 minutes from the time you posted the first time to the second time is too little time to give an old guy like me time to type in the answer.

a) The mass of water can be calculated by finding the difference in mass before and after the water is added to the beaker.

Mass of water = (Mass of beaker + water) - Mass of beaker
Mass of water = 69.122 g - 49.135 g
Mass of water = 19.987 g

b) The actual volume of water in the beaker can be determined using the density formula:

Actual volume of water = Mass of water / Density of water
Actual volume of water = 19.987 g / 0.99732 g/mL
Actual volume of water = 20.045 mL

c) The percent error in the volume measurement can be calculated using the formula:

Percent error = |(Actual value - Theoretical value) / Theoretical value| * 100%
Percent error = |(20.045 mL - 20.00 mL) / 20.00 mL| * 100%
Percent error ≈ 0.225%

So, the mass of water is 19.987 g, the actual volume of water in the beaker is 20.045 mL, and the percent error in the volume measurement is approximately 0.225%.

To answer these questions, we'll need to use the concept of density. Density is defined as the mass of a substance per unit volume. In this case, water has a density of 0.99732 g/mL at 24 degrees Celsius.

Let's go step by step to find the answers:

a) The mass of water:
To determine the mass of water, we need to subtract the mass of the beaker from the combined mass of water and beaker. From the information provided, the mass of the beaker is 49.135 g and the combined mass of water and beaker is 69.122 g.

Mass of water = Combined mass of water and beaker - Mass of beaker
Mass of water = 69.122 g - 49.135 g
Mass of water = 19.987 g (rounded to three decimal places)

b) The actual volume of water in the beaker:
To find the actual volume of water in the beaker, we can use the density of water. We know that the observed volume of water in the volumetric pipet is 20.00 mL.

Actual volume of water in the beaker = Volume of water in the pipet
Actual volume of water in the beaker = 20.00 mL (given)

c) Percent error in the volume measurement:
To calculate the percent error in the volume measurement, we'll compare the actual volume of water in the beaker to the observed volume of water in the volumetric pipet.

Percent error = [(Actual volume - Observed volume)/Observed volume] * 100

Percent error = [(20.00 mL - 20.00 mL)/20.00 mL] * 100
Percent error = (0 mL / 20.00 mL) * 100
Percent error = 0%

Therefore, the percent error in the volume measurement is 0%. This means that the observed volume and the actual volume of water in the beaker are the same.

To summarize:
a) The mass of water is 19.987 g
b) The actual volume of water in the beaker is 20.00 mL
c) The percent error in the volume measurement is 0%

Thank you so much for your help.

I actually sent the second one because I realized I had forgotten to say please in the first one.

69.122 g = mass bkr + mass H2O

-49.135 = mass empty bkr
-----------
a. xxxxxx = mass H2O

b. volume = mass/density = xxxx/0.99732

c. %error = [value of b/20]*100 = ?

The second one was sent while I was typing the response. We're glad to have you use Jiskha. You're welcome to use our services anytime. I would have prefered that yu tell us what you didn't understand about the problem AND to tell us how you thought the problem might be worked.