What is the molar quantity of a 458*g458⋅g mass of "lead(II) sulfide"lead(II) sulfide?

1 Answer
anor277
Jun 1, 2017

Approx. 2*mol2⋅mol.

Explanation:

The molar mass of "lead sulfide"lead sulfide is 239.3*g*mol^-1239.3⋅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)=Moles of lead sulfide=MassMolar mass=458⋅g239.3⋅g⋅mol−1=

1.91*mol1.91⋅mol.

This is dimensionally consistent because (458*g)/(239.3*g*mol^-1)=458⋅g239.3⋅g⋅mol−1=

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

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