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Answer 1 · 2015-12-14T22:24:02

#0,0017mol#

Since concentration is moles per unit volume (#c=n/V#), and moles is mass over molar mass (#n=m/(M_r#), we may combine the two to obtain that

#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#.