General Knowledge - Chemistry - Discussion

Discussion Forum : Chemistry - Section 1 (Q.No. 28)
28.
The number of water molecules present in a drop of water (volume 0.0018 ml) at room temperature is
1.568 x 103
6.023 x 1019
4.84 x 1017
6.023 x 1023
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 2 of 2.

Vikas jha said:   8 years ago
Here, we must have to deal with volume rather than density then the answer will be different. We have to take the volume of one mole of water (22400ml).

Hari said:   9 years ago
Here, d = m/v equal to 0.0018 g so we can write as 18 * 10^-4 g.
Then n = 18 *10^-4/18 = 1 * 10^-4.
One mole = 6.022 * 10^23.
1 * 10^-4 mole = x
And the answer is 6.022 * 10^19.

Neelam said:   1 decade ago
Density of water = 1.

So 1 = mass/volume, 1 = mass/0.0018, mass = 0.0018g no. of mole = weight/molecular

weight = 0.0018/18.

No of water molecules = No of mole*Avogadros no.

= 0.0018/18*6.023*10exp23 = 6.023*10exp19.

Ranjeet Yadav said:   1 decade ago
Number of mole = Molecule.wt*Avogadros no. Thus by multiplying both we get answer.

Deepika said:   1 decade ago
In the question the volume of water is given. To relate this we must have to used volume of one mole equals to 22.4 lit rather than gram molecular mass.

Abhijeet said:   1 decade ago
Why should taken (6.02 x 10exp19 molecules) molecule, if Avogadro.No is 6.023 x 1023?

Can you explain?

Subhash jha said:   1 decade ago
First using the density formula, convert volume to mass then divide the mass substance from molecular weight of compound.

You will get the answer for any molecules.


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