1.

The number of silver atoms present in a 90% pure silver wire weighing 10 g is :

Answer»

`5.57 xx 10^22`
`0.62 xx 10^23`
`5.0 xx 10^22`
`6.2 xx 10^29`

Solution :Amount of PURE SILVER in 10g wire
` = 10 xx 90/100 = 9g `
108 g of Ag contain ATOMS =`6 xx 10^23`
9g of Ag contain atoms
` = (6 xx 10^23)/(108) xx 9= 5 xx 10^22`


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