On your wedding day your lover gives you a gold ring of mass 3.78 g. 50 years later its mass is...

Question:

On your wedding day your lover gives you a gold ring of mass {eq}3.78\ g {/eq}. 50 years later its mass is {eq}3.16\ g {/eq}. On the average how many atoms were abrated from the ring during each second of your marriage?

Gold atom:

For one atom of gold, the mass is

{eq}m = 3.27\times 10^{-22}\ g {/eq}

The molar mass of gold is {eq}196.666\dfrac{g}{mol} {/eq}

Answer and Explanation:

The initial mass of gold, {eq}m_i = 3.78\ g {/eq}

final mass after 50 years, {eq}m_f = 3.16\ g {/eq}

Number of seconds in 50 years is = {eq}50\times 365\times 24\times 60\times 60 = 1.5768\times 10^9\ s {/eq}

Mass of gold lost during 50 years is = {eq}3.78 - 3.16 =0.62\ g {/eq}

Number of atoms lost in 50 years is = {eq}\dfrac{0.62}{ 3.27\times 10^{-22}} {/eq}

Hence, number of atoms of gold lost in 1 sec is ={eq}\dfrac{0.62}{ 3.27\times 10^{-22}\times 1.5768\times 10^9} = 1.2\times 10^{12} {/eq}


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