Question 763409
(a)
Let {{{ a }}} = ml of solution 1 to be added
{{{ .8a }}} = ml of benzene in solution 1
{{{ .3*500 = 150 }}} ml of benzene in solution 2
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{{{ ( .8a + 150 ) / ( a + 500 ) = .7 }}}
{{{ .8a + 150 = .7*( a + 500 ) }}}
{{{ .8a + 150 = .7a + 350 }}}
{{{ .1a = 200 }}}
{{{ a = 2000 }}}
2000 ml ( 2 l ) of solution 1 must be added
check:
{{{ ( .8a + 150 ) / ( a + 500 ) = .7 }}}
{{{ ( .8*2000 + 150 ) / ( 2000 + 500 ) = .7 }}}
{{{ ( 1600 + 150 ) / 2500 = .7 }}}
{{{ 1750 = .7*2500 }}}
{{{ 1750 = 1750 }}}
OK
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(b)
Let {{{ a }}} = ml of solution 1 needed
Let {{{ b }}} = ml of solution 2 needed
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(1) {{{ a + b = 100 }}}
(2) {{{ ( .8a + .3b ) / 100 = .5 }}}
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(2) {{{ .8a + .3b = .5*100 }}}
(2) {{{ .8a + .3b = 50 }}}
(2) {{{ 8a + 3b = 500 }}}
Multiply both sides of (1) by {{{ 3 }}}
and subtract (1) from (2)
(2) {{{ 8a + 3b = 500 }}}
(1) {{{ -3a - 3b = -300 }}}
{{{ 5a = 200 }}}
{{{ a = 40 }}}
and
(1) {{{ a + b = 100 }}}
{{{ b = 60 }}}
40 ml of solution 1 is needed
60 ml of solution 2 is needed
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(c)
No. You can't get a greater % of benzene than
either of the 2 solutions have which is
80% and 30%