In this question, the density of cobalt is equal to 8.86 g per cubic centimeter. The mass of average atom of cobalt is equal to 9.786 316 times 10. Raise to part minus 23 grand. So this is the mass of average item of cobalt. And we have to calculate the number of atoms of cobalt, when volume is equal to one cc. So in one cc of cobalt we have to calculate the number of atoms of COBOL present. First of all, using the formula that density is equal to mass divided by volume. We can calculate the mass of atoms present in one cc. So the mass would be equal to density, multiplied by volume. From this mass comes out to be density is 8.86 g per centimeter cube and volume is one centimeter tube. So the mass would be equal to 8.86 g. Now we have the mass of atoms present in one cc of COBOL. Now, since mass is 8.86 g and we know that 9.7863 16 times 10 raise to part -23 g is equal and to one atom since it is a mass of average atom of cobalt. So from this we can say that 8.86 g of cobalt would be equal to one divided by 9.7686316 multiply by 8.86 atoms of Cobalt. From this. By solving it, we will get the value which is equal to when we multiply and divide these things. We will get the value 9.5 times 10, raise to part 22 atoms, so 9.5 Multiply by 10 days to part 22 items are present in 8.86 g of cobalt. Or we can say that one cc of cobalt. So our answer is option A.
I hope this helps! o(〃＾▽＾〃)o