Question #bafc0

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Question #bafc0

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Answer 1 · 2015-12-14T22:24:02

$0, 0017 m o l$ $0,0017mol$

Explanation:

Since concentration is moles per unit volume ( $c = \frac{n}{V}$ $c=n/V$ ), and moles is mass over molar mass ( $n = \frac{m}{M_{r}}$ $n=m/(M_r$ ), we may combine the two to obtain that

$c = \frac{m}{M_{r} \cdot V}$ $c=m/(M_r*V)$

$therefore m=c*M_r*V$

$=2,28mol//dm^3xx(39+80)g//molxx0,00075dm^3$

$=0,2035g$ .

This corresponds to the following number of moles :

$n=m/(M_r)=(0,2035g)/(119g//mol)=0,0017mol$ .

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