Question

Given,1 GAM sodium means 23 g sodium. This contains $\text{ 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ }$atoms. If so, how many GAM is present in the 69 g of sodium? How many atoms are present in it?

Answer 1

Here the question can be solved based on data provided in the problem. The 1 GAM is equal to the 23 g of sodium and 23 g of sodium contains the $\text{ 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ }$atoms .Therefore the 1 GAM of sodium contains the $\text{ 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ }$of atoms. The relation can be written as,
$\text{ 1 GAM of Na = 23 g of Na = 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms }$

Complete answer:
We have given the following data:
1 GAM sodium is equal to the 23 g of sodium
The 1 GAM contains the $\text{ 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ }$ atoms.
We are interested to find out the number of GAM present in the 69 g of sodium.
Let’s first write down the relation, 1 GAM sodium equal to 23 g of sodium that is,
$\text{ 1 GAM of Na = 23 g of Na }$
Then 69 grams of the sodium will contain an ‘x’ number of GAM in it.
$\text{ 69 g of Na = }\\!\\!'\\!\\!\text{ x }\\!\\!'\\!\\!\text{ GAM of Na}$
On cross multiplying The value of ‘x’ is then equal to,
$\text{ X GAM of Na = }\dfrac{69}{23}\text{ }\times \text{ 1 GAM of Na = 3 GAM of Na }$
Thus,69 grams of sodium contains 3 GAM of sodium in it.
We have given that, The 1 GAM or the 23 g of sodium contains $\text{ 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ }$atoms. That is,
$\text{ 1 GAM of Na = 23 g of Na = 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms }$
Therefore, the number of atoms present in the 69 gram of sodium or 3 GAM of sodium is given by,
$\text{ No}\text{.of atoms = 3 }\times \text{ 6}\text{.22 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms = 1}\text{.866 }\times \text{ 1}{{\text{0}}^{\text{24}}}\text{ atoms }$
Thus, 69 grams of sodium contains the $\text{ 1}\text{.866 }\times \text{ 1}{{\text{0}}^{\text{24}}}\text{ }$ atoms in it.
Thus 69 grams of sodium contains,
i) 3 GAM of sodium
ii) $\text{ 1}\text{.866 }\times \text{ 1}{{\text{0}}^{\text{24}}}\text{ }$Of atom in it.

Note:
Note that, this question is pretty straightforward. If A is equal to B and B is equal to C, then in turn A is equal to the C. $\text{ A= B }$ And $\text{ B = C }$ then $\text{ A=C }$ .
First established as the relation between the 1 GAM of sodium with sodium and next is the relation of sodium with the number of atoms and we will find the answer.