from graph ,at particular temparature the mass of salt lying on
the line can be obtained.That
mass is the amount required for saturated solution.Points on line
represent
saturaed solution Below that value will be
unsaturated,above the vale will be super saturated
A) 20 gof NaCl at 30 C = unsaturated
b) 50 g KCl at 50 C= super saturated
c)10 g KCLO3 50 g water at 40 C= unsaturated
d) 80 g KNO3 = super saturated
e) 80 g Pb(NO3)2 = unsaturated
****************
A)
moles of NaCl = 2 mols
volume of water = 400 ml = 0.4 L
molarity = moles of NaCl[solute]/volume of water[solution]
= 2 mols/).4 L = 5 M
***********************
b)
mass of KBr = 10g
molar mass of KBr = [1*39.10 + 1*79.9] g/mol= 119.0 g/mol
moles of KBr present = 10 g/119 g/mol=0.084 mols
volume of solution = 200 ml = 0.2 L
molarity = number of moles of KBr/volume of solution = 0.084
mols/0.2 L
= 0.42 M
***************
c) given molarity = 154 milli M = 154 x1o^-3 M
volume = 3L
so moles of NaCl required = molarity*volume
=154 x1o^-3 M*3L = 0.462 M
Mass of NaCl needed = moles of NaCl needed*molar mass of NaCl=
0.462 mols*58.44 g/mol
= 26.999 g
********************
d)
given mass of solute AgNO3 = 50 g
volume of solution = 600 ml
%[m/v] = mass of solute/volume of solution in ml *100
= 50 /600**100 = 8.333 $
***********
5[v/v] = 4.2 means
volume of alcohol = 4.2 ml =0.142 ounce
volume of beer = 100 ml = 0.211 pints
SO in 0.211 pints there is 0.142 ounce
So in 1 pint = 0.142/0.211 = 0.67 ounce
************************
All solved .Kindly upvote :)
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