1.

Calculate the number of milliequivalents, gram eq., weight in grams and number of moles contained in 10 litres of 0.5 M Na(OH)_(2) solution. (Ba=137) .

Answer»

Solution :Normality of `Ba(OH)_(2)` solution = MOLARITY `xx`acidity …(Eqn.6)
`=0.5 xx2`
1 N
m.e of the solution = normality `xx` vol.in mL
` = 1 xx 10000`
`=10000`
Equivalent of the solution = `(m.e)/(1000)""…(Eqn.3)`
`=(1000)/1000`
=10
Weight of `Ba(OH)_(2)` in solution = equivalent `xx` eq.wt ....(Eqn.4)
`= 10 xx85.5`
= 855 G
`{"equivalent wt. of "Ba(OH)_(2)=("mol.wt")/("acidity")xx171/2 = 85.5}`
Moles of `NA(OH)_(2)` solution = `("wt. in grams")/("mol .wt")`
`= (855)/171 = 5.00 ` moles .


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