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:
Search up Cell Divison Reinforcement key and it pops up
Explanation:
Literally took 30 seconds to find online or individually ask every question.
Answer:
Ba(OH)2(aq)+H2SO4(aq) gives us 2BaH+H2O
Explanation:
Answer:
Option b) polyatomic ions
Explanation:
Polyatomic ions are ions consisting of two or more atoms.
From the question given above, we can see that each ions consist of more than one atom as shown below:
Ions >>>>>> Number of atom present
NH4+ >>>>> 2
CO32− >>>> 2
PO43− >>>> 2
Thus, we can say that the above ions are polyatomic ions.
Answer: C) 0.020 m
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
where,
n = moles of solute
= weight of solvent in g
Mole fraction of 
is = 
i.e. moles of is present in 1 mole of solution.
Moles of solute 
=
moles of solvent (water) = 1 - = 0.99
weight of solvent =
Molality =
Thus approximate molality of in this solution is 0.020 m
