Answer:
0.0365
Explanation:
Three significant figures after the decimal include all numbers except the zeros before other numbers.
For example 0.000003431232 to three significant figures would be 0.00000343. However if the numbers are before the decimal, the zeros are included, for example 19828300 to three significant numbers would be 19800000.
a) Volume percent
Formula: % v/v = [volume solute / volume solution] * 100
Just to make it easy take a base of 50 volume parts of ethylen glycol and 50 volume parts of water to make 100 volumes of mixture (this assumpion will be valid for all the questions):
% v/v =[ 50 ml ethyleneglycol] / [100 ml mixture] * 100 = 50%
Answer: 50% v/v
b) Mass percent
% m/m = [mass ethylene glycol / mass solution] * 100
mass ethylene glycol = 50 ml * 1.114 g/ ml = 55.7 g
mass of mixture = 100 ml * 1.07 g/ml = 107 g
% m/m = [55.7 / 107 g] * 100 = 52.06 %
Answer: 52.06%
c) Molarity
M = number of moles of solute / liters of solution
number of moles of solute = mass in grams / molar mass
number of moles of ehtylene glycol = 55.7 g / 62.07 g/mol = 0.8974 mol
liters of solution = 0.1 liter
M = 0.8974 mol / 0.1 liter = 8.974 M
Answer: 8.974 M
d) Molality
m = number of moles of solute / kg of solvent
number of moles of ethylen glycol = 0.8974 mol
mass of water = 50 ml * 1 g/ml = 50 g = 0.05 kg
m = 0.8974 mol / 0.05 kg = 17.95 m
Answer: 17.95 m
e) mole fraction
mole fraction = [number of moles of solute] / [number of moles of mixture] * 100
number of moles of ethylen glycol = 0.8974 mol
number of moles of water = 50 g / 18.01 g /mol = 2.776 mol
mole fraction = 0.8974 mol / [0.8974 mol + 2.776 mol] = 0.244
Answer: 0.244
Answer:
.
Explanation:
Magnesium chloride and silver nitrate reacts at a 
ratio:
.
In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:
.
The precipitate silver chloride 
is insoluble in water and barely ionizes. Hence, 
isn't rewritten as ions.
Net ionic equation:
.
Calculate the initial quantity of nitrate ions in the mixture.
.
Since nitrate ions 
do not take part in any reaction in this mixture, the quantity of this ion would stay the same.
.
However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be 
of the concentration in the original solution.
.
