A glucose solution is prepared by dissolving 5.10 g of glucose, C6H12O6, in 110.5 g of water. What is the molality of the glucose solution?
A) 0.283 m B) 0.000256 m C) 0.245 m D) 0.256 m
2006-12-12 05:31:28 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Molality is the number of moles in 1000g of SOLUTION (i.e. solute + solvent).
So number of mol of glucose = 5.10 / 180 = 0.0283 mol
These are present in 5.10g + 110.5g of solution = 115.6g
So 0.0283 mol in 115.6g, ? in 1000g
0.0283*1000/115.6 = 0.245m
2006-12-12 05:42:28 · answer #1 · answered by claudeaf 3 · 0⤊ 1⤋
The molality concentration is given as:
molality = moles of solute / kilograms of solvent
You know the mass of the solute (glucose), you need to convert this to moles using the molar mass of glucose.
You are given the mass of the solvent (water), you need to convert to kg by dividing by 1000 g/kg.
The molar mass of glucose is: 180.16 g/mol
5.10 grams / 180.16 g/mol = .0283 moles of glucose
molality = .0283 moles / .1105 kg
molality = .256 = choice D)
http://dbhs.wvusd.k12.ca.us/webdocs/Solutions/Molality.html
2006-12-12 05:39:50 · answer #2 · answered by mrjeffy321 7 · 0⤊ 0⤋
molality=mole of solute/ kg of solvent
mol of glucose=5.1g x 1mol/180.08g =.0283mol
molality=.0283mol/.1105kg= .256m
The answer is d
2006-12-12 06:27:05 · answer #3 · answered by lynn 2 · 0⤊ 0⤋
C is the answer indeed
:> peace
.
2006-12-12 09:02:28 · answer #4 · answered by Anonymous · 2⤊ 1⤋
C
2006-12-12 05:41:36 · answer #5 · answered by pkababa 4 · 0⤊ 5⤋