Please help me solve this problem & show work. I have no idea how to solve it
Calculate the percent by mass of the solute in a solution made of 4.5 g of NaCl in 95.6 mL of water at 20 degrees C (celcius). If the same solution was prepared at 25 degrees C (celcius) what is the mass percent? Is there a significant difference between these two mass percents?
CHEMISTRY percent by mass of solute?
2015-05-26 5:43 pm
回答 (3)
2015-05-26 6:10 pm
For 20 C :
-------------------
d H2O = 0.998 g / mL
m H2O = ( d H20 ) ( V H2O )
m H2O = ( 0.998 g / mL ) ( 95.6 mL ) = 95.4 g H2O
mass % NaCl = ( 4.5 g NaCl / 4.5 g NaCl + 95.4 g H2O )( 100 )
mass % NaCl = 4.5 mass % NaCl <--------------------
You have for 25 C :
---------------------------------
d H2O = 0.998 g / mL
m H2O = ( d H2O ) ( V H2O ) = ( 0.998 g / mL ) ( 95.6 mL ) = 95.4 g H2O
mass % NaCl = ( 4.5 g NaCL / 4.5 g NaCl + 95.4 g H2O ) ( 100 )
mass % NaCl = 4.50 mass % NaCl <-------------------
Calculations for 20 C and for 25 C give the same values for the masses
of water and mass percent NaCl [ 4.50 mass % NaCl ] <------------------------
-------------------
d H2O = 0.998 g / mL
m H2O = ( d H20 ) ( V H2O )
m H2O = ( 0.998 g / mL ) ( 95.6 mL ) = 95.4 g H2O
mass % NaCl = ( 4.5 g NaCl / 4.5 g NaCl + 95.4 g H2O )( 100 )
mass % NaCl = 4.5 mass % NaCl <--------------------
You have for 25 C :
---------------------------------
d H2O = 0.998 g / mL
m H2O = ( d H2O ) ( V H2O ) = ( 0.998 g / mL ) ( 95.6 mL ) = 95.4 g H2O
mass % NaCl = ( 4.5 g NaCL / 4.5 g NaCl + 95.4 g H2O ) ( 100 )
mass % NaCl = 4.50 mass % NaCl <-------------------
Calculations for 20 C and for 25 C give the same values for the masses
of water and mass percent NaCl [ 4.50 mass % NaCl ] <------------------------
2015-05-26 6:03 pm
At 20C, the density of water is 0.998 g/mL. So, the mass of water at this temperature is:
95.6 mL X 0.998 g/mL = 95.4 g. The %solute (m/m) = (4.5 g / 99.9) X 100 = 4.504%
At 25C, the density of water is 0.997 g/mL. So, the mass of water at this temperature is:
95.6 mL X 0.997 g/mL = 95.3 g. The %solute (m/m) = (4.5 g / 99.8) X 100 = 4.509%
Because your final answer can only have 2 significant figures (because of the 4.5 g solute), there is no significant difference between the two temperatures.
95.6 mL X 0.998 g/mL = 95.4 g. The %solute (m/m) = (4.5 g / 99.9) X 100 = 4.504%
At 25C, the density of water is 0.997 g/mL. So, the mass of water at this temperature is:
95.6 mL X 0.997 g/mL = 95.3 g. The %solute (m/m) = (4.5 g / 99.8) X 100 = 4.509%
Because your final answer can only have 2 significant figures (because of the 4.5 g solute), there is no significant difference between the two temperatures.
2015-05-26 6:00 pm
Mass % is 4.4/100 x 100% either way = 4.5%
收錄日期: 2021-05-01 20:35:25
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