Question #bb39f

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Answer 1 · 2016-10-11T05:38:06

Molar mass of
$CuNO_3=(63.5+14+3*16)=125.5" g/mol"$

The amount of $CuNO_3=6.58g=(6.58g)/(125.5g/"mol")$
$=0.052mol$

If the volume of the solution be VL,then its strength $=0.052/VM$

Given strength is 0.15M

By the problem

$0.052/V=0.15$

$=>V=0.052/0.15L=0.3467L=346.7L$