50.0 ml of methanol (density = 0.791 g/ml, MM = 32.05 g/mol) and 50.0 ml of water (density = 1.00 g/ml, MM = 18.02 g/ml) are mixed in a beaker and make a solution with a final volume of 96.0 ml.
A) What is the density of the new solution?
b) What is the %w/v of the new solution?
c) W hat is the %w/w of the new solution?
d) What is the molar concentration of the new solution?
help please...................!!!!!!!!!!?
2020-08-12 7:40 am
回答 (2)
2020-08-12 9:23 am
A)
Mass of the new solution = (50.0 × 0.791 + 50.0 × 1.00) g = 89.55 g
Volume of the new solution = 96.0 ml
Density of the new solution = 89.55/96.0 g/ml = 0.933 g/ml
====
B)
Methanol and water are miscible. Suppose that methanol is the solute and water is the solvent.
Mass of methanol = 50.0 × 0.791 g = 39.55 g
%w/v of the new solution = (39.55/96.0) × 100%w/v = 41.2%w/v
====
C)
%w/w of the new solution = (39.55/89.55) × 100%w/w = 44.2%w/w
====
D)
Moles of methanol = (39.55 g) / (32.05 g/mol) = 1.234 mol
Volume of the new solution = 96.0 ml = 0.0960 L
Molar concentration of the new solution = 1.234/0.0960 M = 12.9 M
Mass of the new solution = (50.0 × 0.791 + 50.0 × 1.00) g = 89.55 g
Volume of the new solution = 96.0 ml
Density of the new solution = 89.55/96.0 g/ml = 0.933 g/ml
====
B)
Methanol and water are miscible. Suppose that methanol is the solute and water is the solvent.
Mass of methanol = 50.0 × 0.791 g = 39.55 g
%w/v of the new solution = (39.55/96.0) × 100%w/v = 41.2%w/v
====
C)
%w/w of the new solution = (39.55/89.55) × 100%w/w = 44.2%w/w
====
D)
Moles of methanol = (39.55 g) / (32.05 g/mol) = 1.234 mol
Volume of the new solution = 96.0 ml = 0.0960 L
Molar concentration of the new solution = 1.234/0.0960 M = 12.9 M
2020-08-12 7:55 am
move to homework help. lazy worthless pos cheater. What makes you think you are learning anything by having others do your work?
收錄日期: 2021-04-24 07:57:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200811234010AAjt3Fd