What is the molar quantity of a #458*g# mass of #"lead(II) sulfide"#?

1 Answer
Jun 1, 2017

Answer:

Approx. #2*mol#.

Explanation:

The molar mass of #"lead sulfide"# is #239.3*g*mol^-1#.

And to get the molar quantity of the given mass of lead sulfide, we take the quotient.

#"Moles of lead sulfide"="Mass"/"Molar mass"=(458*g)/(239.3*g*mol^-1)=#

#1.91*mol#.

This is dimensionally consistent because #(458*g)/(239.3*g*mol^-1)=#

#cancelg/(cancelg*mol^-1)=1/(1/(mol))=mol# as required..............