R RING12 Member 10+ Year Member 15+ Year Member Joined Aug 13, 2006 Messages 207 Reaction score 0 Points 0 Pre-Dental Mar 6, 2011 #1 Advertisement - Members don't see this ad A researcher adds 0.4 kg of CaBr2 (MW=200g/mol) to 10L of water, in which it disolves completely: How much water would the researcher need to add to the solution in order to decrease the concentration by factor of 4?😱
Advertisement - Members don't see this ad A researcher adds 0.4 kg of CaBr2 (MW=200g/mol) to 10L of water, in which it disolves completely: How much water would the researcher need to add to the solution in order to decrease the concentration by factor of 4?😱
W wizi Full Member 15+ Year Member Joined Aug 24, 2008 Messages 960 Reaction score 9 Points 4,551 Dental Student Mar 6, 2011 #2 RING12 said: A researcher adds 0.4 kg of CaBr2 (MW=200g/mol) to 10L of water, in which it disolves completely: How much water would the researcher need to add to the solution in order to decrease the concentration by factor of 4?😱 Click to expand... lol Lets me try... 1st: lets define some variables. x: amount of water we are looking for. C2= new molality. C1 = old molality. --> C2 = C1/4 2nd: Find C1 C1 = mole (CaBr2)/ kg of water mole of CaBr2 = 400/200 = 2 moles. kg of water = 10kg (water has density of 1 --> Litter = kg) --> C1 = 2/10 = 0.2 3rd: Find C2 C2= C1/4 = 0.2/4 = 0.05 4th: Find x C2 = mole (CaBr2)/ x(kg) --> x (kg) = C2 / mole (CaBr2) = 0.05 / 2 = 0.025 (kg) = 25 (g) Answer: 25 (g) Is it right? Upvote 0 Downvote
RING12 said: A researcher adds 0.4 kg of CaBr2 (MW=200g/mol) to 10L of water, in which it disolves completely: How much water would the researcher need to add to the solution in order to decrease the concentration by factor of 4?😱 Click to expand... lol Lets me try... 1st: lets define some variables. x: amount of water we are looking for. C2= new molality. C1 = old molality. --> C2 = C1/4 2nd: Find C1 C1 = mole (CaBr2)/ kg of water mole of CaBr2 = 400/200 = 2 moles. kg of water = 10kg (water has density of 1 --> Litter = kg) --> C1 = 2/10 = 0.2 3rd: Find C2 C2= C1/4 = 0.2/4 = 0.05 4th: Find x C2 = mole (CaBr2)/ x(kg) --> x (kg) = C2 / mole (CaBr2) = 0.05 / 2 = 0.025 (kg) = 25 (g) Answer: 25 (g) Is it right?
J jkb53433 Full Member 10+ Year Member Joined Jan 2, 2011 Messages 201 Reaction score 0 Points 1 Pre-Dental Mar 7, 2011 #3 whats the answer? im getting 30L but i dunno if im on the right track Upvote 0 Downvote
Q qkchen Full Member 10+ Year Member Joined Nov 17, 2009 Messages 266 Reaction score 0 Points 0 Dental Student Mar 7, 2011 #4 jkb53433 said: whats the answer? im getting 30L but i dunno if im on the right track Click to expand... i got 30 ml of water to add, as well, with V2 = 40 ml Upvote 0 Downvote
jkb53433 said: whats the answer? im getting 30L but i dunno if im on the right track Click to expand... i got 30 ml of water to add, as well, with V2 = 40 ml
R RING12 Member 10+ Year Member 15+ Year Member Joined Aug 13, 2006 Messages 207 Reaction score 0 Points 0 Pre-Dental Mar 7, 2011 #5 the right answer is 30L. thanks every one Last edited: Mar 7, 2011 Upvote 0 Downvote
