How many molecules are in 3.6 grams of NaCl? Question options: A 0.06
B 1.0 X 10^21
C 1.3 x 10^26
D 3.7 x 10^22

Explanation:

We are asked to find how many molecules are in 3.6 grams of sodium chloride.

First, we convert grams to moles using the molar mass. These values are equivalent to atomic masses on the Periodic Table, but the units are grams per moles instead of atomic mass units. Look up the molar masses of the individual elements: sodium and chlorine.

There are no subscripts in the chemical formula (NaCl), so we simply add the 2 molar masses.

Now we will convert using dimensional analysis. First, set up a ratio using the molar mass.

We are converting 3.6 grams to moles, so we must multiply the ratio by this value.

Flip the ratio so the units of grams of sodium chloride cancel.

Next, we convert moles to molecules using Avogadro's Number. This is 6.022 × 10²³ and it tells us the number of particles (atoms, molecules, formula units, etc). In this case, the particles are molecules of sodium chloride. Let's set up another ratio.

Multiply by the number of moles we calculated.

The units of moles of sodium chloride cancel.

The original measurement of grams (3.6) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredth place tells us to leave the 7 in the tenth place.

There are in 3.6 grams and the correct answer is choice D.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole