The total number of protons in 10 g of calcium carbonate is `(N_(0) = 6.023 xx 10^(23))` :-

Options

A. `1.5057 xx 10^(24)`
B. `2.0478 xx 10^(24)`
C. `3.0115 xx 10^(24)`
D. `4.0956 xx 10^(24)`

Answer

C

Explanation

` because ` 100g `CaCO_(3) = 6.022 xx 10^(23)` molecules
`therefore` 10 g `CaCO_(3) = (6.022 xx 10^(23))/(100) xx 10`
` = 6.022 xx 10^(22)` molecule
1 molecule of `CaCO_(3)` = 50 protons
`6.022 xx 10^(22)` molecule of `CaCO_(3) = 50 xx 6.022 xx 10^(22)`
` = 3.0115 xx 10^(24)`