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A. $\dfrac{1.2}{35}\times {{\text{N}}_{\text{A}}}$

B. $\dfrac{1}{35}\times {{\text{N}}_{\text{A}}}$

C. $\dfrac{1.2}{{{\left( 35 \right)}^{2}}}\times {{\text{N}}_{\text{A}}}$

D. $1.2\text{ }{{\text{N}}_{\text{A}}}$

Answer
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Let us solve this question step by step:

Step (1)- The density of liquid is 1.2 g/mL.

It means that 1 mL has 1.2 grams.

Then, 2 mL of liquid will have $1.2\times 2$ or 2.4 grams.

So, it means that 2.4 grams of liquid will have 35 drops or 35 drops will make up 2.4 grams of liquid.

Step (2)- We know that 1 mole of liquid will have ${{\text{N}}_{\text{A}}}$ molecules.

1 mole means molar mass of the liquid.

Molar mass of liquid is 70 grams.

So, 70 grams of liquid will have ${{\text{N}}_{\text{A}}}$ molecules.

Step (3)- The number of drops in 70 grams of liquid will be calculated by unitary method;

As, we already know, 2 = 2.5 gm of the liquid.

So, if 2.4 gm has 35 drops, then,

$\dfrac{70}{2.4}\times 35$ = 1020.8 drops.

That is 1021 drops (approximately) are present in 70 grams.

There are ${{\text{N}}_{\text{A}}}$ molecules in 1021 drops,

So the number of molecules present in one drop will be $\dfrac{{{\text{N}}_{\text{A}}}}{1021}\text{ or }\dfrac{{{\text{N}}_{\text{A}}}\times 2.4}{70\times 35}$.

As, 1021 has the value $\dfrac{70\times 35}{2.4}$.

The molecules in drop are $\dfrac{1.2}{35\times 35}\times {{\text{N}}_{\text{A}}}$ or $\dfrac{1.2}{{{\left( \text{35} \right)}^{2}}}\times {{\text{N}}_{\text{A}}}$.

The correct option is option ‘c’ which is the number of molecules of liquid in one drop are $\dfrac{1.2}{{{\left( \text{35} \right)}^{2}}}\times {{\text{N}}_{\text{A}}}$.

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