The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70)

Question

The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70) (A) (1.2/35)NA (B) ((1/35))2 NA (C) (1.2/(35)2)NA (D) 1.2NA

Tardigrade Admin · Accepted Answer

Volume of one drop =(2/35)mL Number of moles in one drop =(2×1.2/35×70)=(1.2/(35)2) Number of molecules in one drop =(1.2/(35)2)× NA (NA = Avogadroâs number)

Q. The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70)

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

$(\frac{1}{35})^{2} N_{A}$

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

$1.2 N_{A}$

Volume of one drop $= \frac{2}{35} m L$
Number of moles in one drop $= \frac{2 \times 1.2}{35 \times 70} = \frac{1.2}{( 35 ) ^{2}}$
Number of molecules in one drop $= \frac{1.2}{( 35 ) ^{2}} \times N_{A}$
( $N_{A} =$ Avogadro’s number)

Q. The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70)

351.2​NA​

(351​)2NA​

(35)21.2​NA​

1.2NA​

Volume of one drop =352​mL Number of moles in one drop =35×702×1.2​=(35)21.2​ Number of molecules in one drop =(35)21.2​×NA​ (NA​= Avogadro’s number)

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

$(\frac{1}{35})^{2} N_{A}$

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

$1.2 N_{A}$

Volume of one drop $= \frac{2}{35} m L$
Number of moles in one drop $= \frac{2 \times 1.2}{35 \times 70} = \frac{1.2}{( 35 ) ^{2}}$
Number of molecules in one drop $= \frac{1.2}{( 35 ) ^{2}} \times N_{A}$
( $N_{A} =$ Avogadro’s number)