The density of liquid (mol. wt. = 70) is 1.2 g $m L^{- 1}$ . If 2 mL of liquid contains 35 drops, the number of molecules of liquid in one drop are :

\frac{1.2 \times N_{A}}{3.5}

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\frac{1.2 \times N_{A}}{35}

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\frac{1.2 \times N_{A}}{(35)^{2}}

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1.2 N_{A}

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Solution

The correct option is C $\frac{1.2 \times N_{A}}{(35)^{2}}$
wt. of one drop of liquid = $\frac{(1.2 \times 2)}{35}$ gram
70 gram of liquid contain NA molecule
$\frac{(1.2 \times 2)}{35}$ gram of liquid contain = $\frac{(N_{A} \times 1.2 \times 2)}{(70 \times 35)} = \frac{1.2}{(35)^{2}} N_{A}$

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The density of a liquid (mol. wt. = 70) is 1.2 g $m L^{- 1}$ . If 2 mL of liquid contains 35 drops, the number of molecules of liquid in one drop are :

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Q. The density of a liquid (molar mass 70) is 1.2g/mL. If 2 mL of liquid contains 35 drops,the number of molecules of liquid in one drop is :

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The density of liquid (mol. wt. = 70) is 1.2 g mL−1. If 2 mL of liquid contains 35 drops, the number of molecules of liquid in one drop are :

The correct option is C 1.2×NA(35)2wt. of one drop of liquid = (1.2×2)35 gram 70 gram of liquid contain NA molecule (1.2×2)35 gram of liquid contain = (NA×1.2×2)(70×35)=1.2(35)2NA

The density of liquid (mol. wt. = 70) is 1.2 g $m L^{- 1}$ . If 2 mL of liquid contains 35 drops, the number of molecules of liquid in one drop are :

The correct option is C $\frac{1.2 \times N_{A}}{(35)^{2}}$
wt. of one drop of liquid = $\frac{(1.2 \times 2)}{35}$ gram
70 gram of liquid contain NA molecule
$\frac{(1.2 \times 2)}{35}$ gram of liquid contain = $\frac{(N_{A} \times 1.2 \times 2)}{(70 \times 35)} = \frac{1.2}{(35)^{2}} N_{A}$