Please don't tell me the whole "I can help you but giving you the answer does nothing for you" comment because to be honest I don't care. I'm 2 weeks away from graduating and I have a 76 C in that class and that's too close to failing. I'm enlisted in the NAVY so no, I won't be going to college, at least any time soon, so I don't really care nor need to know this stuff.
1. A 12.5-g sample of glucose (C6H12O6) is dissolved in 225 g of water. Calculate the percent by mass of glucose in the solution.
2. A chemist prepares some standard solutions for use in the lab using 500.0-mL volumetric flasks to contain the solutions. If the following masses of solutes are used, calculate the resulting molarity of each solution:
a. 4.865 g NaCL b. 78.91 g AgNO3
3. Suppose that each of the following solutions is diluted by adding the indicated amount of water. Calculate the new concentrations of solutions.
a. 255 mL of 3.02 M HCl; 365 mL water added
b. 75.1 g of 1.51% AgNO3; 125 g water added
Thanks in advance
1. A 12.5-g sample of glucose (C6H12O6) is dissolved in 225 g of water. Calculate the percent by mass of glucose in the solution.
2. A chemist prepares some standard solutions for use in the lab using 500.0-mL volumetric flasks to contain the solutions. If the following masses of solutes are used, calculate the resulting molarity of each solution:
a. 4.865 g NaCL b. 78.91 g AgNO3
3. Suppose that each of the following solutions is diluted by adding the indicated amount of water. Calculate the new concentrations of solutions.
a. 255 mL of 3.02 M HCl; 365 mL water added
b. 75.1 g of 1.51% AgNO3; 125 g water added
Thanks in advance
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Good Luck with your NAVY.Remember chemistry will be useful in NAVY too.
Hope these answers will help you to pass the course.
1. % mass of glucose= [mass of glucose /(mass of glucose + water) ] x100= [12.5/( 12.5 + 225)] x100= 5.26
2. a) Molarity= moles/volume= (4.865g/ 58.5g/mol)/.5000L= 0.1663M
b) M= (78.91g/ 169.9g/mol)/.5000L= 0.9289M
3.a) (255 mLx 3.02M) / (365+ 255)mL= 1.24M
b) (75.1 g x 1.51%)/ (75.1 +125)g = 0.567%
Hope these answers will help you to pass the course.
1. % mass of glucose= [mass of glucose /(mass of glucose + water) ] x100= [12.5/( 12.5 + 225)] x100= 5.26
2. a) Molarity= moles/volume= (4.865g/ 58.5g/mol)/.5000L= 0.1663M
b) M= (78.91g/ 169.9g/mol)/.5000L= 0.9289M
3.a) (255 mLx 3.02M) / (365+ 255)mL= 1.24M
b) (75.1 g x 1.51%)/ (75.1 +125)g = 0.567%