Question

Ordinary water contains one part of heavy water per 6000 parts by weight. The number of heavy water molecules present in drop of water of volume 0.01 ml is (Density of water is 1gm/ml)

$1)2.5 \times {10}^{16}$
$2)2.5 \times {10}^{17}$
$3)5 \times {10}^{16}$
$4)7.5 \times {10}^{16}$


​

Answer 1

Question:- Ordinary water contains one part of heavy water per 6000 parts by weight. The number of heavy water molecules present in drop of water of volume 0.01 ml is (Density of water is 1gm/ml).

Solution:-

Given:-

1 gram of heavy water in 6000 gram of ordinary water.

Volume(V) = 0.01mL

Density($\rho$) = 1g/ml

Thus,

Mass = Volume × Density

Mass = 0.01mL × 1g/mL

Mass = 0.01g

If 6000g of ordinary water contains 1g of heavy water.

Then, 0.01 g of ordinary water will contain

$\sf\dfrac{1g}{6000}$ × 0.01

=> 1.6 × 10^-6 g of heavy water

To calculate number of moles..

(1.6 × 10^-6g)/(20g/mol)

=> 8 × 10^-8 mol

Now, 1 mole of heavy water contains 6.022 × 10^23 molecules of heavy water.

Then,

8×10^-8 moles of heavy water will contain

[(6.022 × 10^23)/1] × 8 × 10^-8

=> 4.8 × 10^16

By rounding off we will get,

approximate 5 × 10^16 molecules.

____________________________________

$1)2.5 \times {10}^{16}$

$2)2.5 \times {10}^{17}$

$3)5 \times {10}^{16}$✔✔

$4)7.5 \times {10}^{16}$

___________________________________

Answer 2

$\huge\bf\pink{\mid{\overline{\underline{Your\:Solution}}}\mid}$

$\huge{\bold{\boxed{\boxed{\mathfrak{\red{QUESTION}}}}}}$

Ordinary water contains one part of heavy water per 6000 parts by weight. The number of heavy water molecules present in drop of water of volume 0.01 ml is (Density of water is 1gm/ml)?

$\huge{\bold{\boxed{\boxed{\mathfrak{\green{ANSWER}}}}}}$

$Ordinary\:Water\:contains\:one\:part\:of\:heavy\:water\:per\:6000\:parts\:by\:weight$

$That\:means\:6000\:g\:of\:Ordinary\:Water\:contains\:1\:g\:of\:heavy\:water$

$The\:Given\:Mass\:of\:Ordinary\:Water\:= Volume\times\:Density$

$=0.01\:\times\:1=0.01\:g.$

$So\:0.01\:g\:of\:Ordinary\:Water\:Contains\:\\=\frac{1}{6000}\:\times\:0.01\\=1.67\:\times\:10^{-6} g\:of\:heavy\:water$

$No\:of\:moles\:of\:heavy\:water=\frac{mass\:of\:the\:heavy\:water}{molar\:mass\:of\:D_{2}O (4+16)  }$

$=\frac{1.67\:\times10^{-6}}{20}$

$=0.0835\times10^{-6}\:mol$

$\bold\:No\:of\:moles=mole\:\times\:avogadro\:number$

$=0.0835\:\times\:10^{-6}\:\times\:6.022\:\times\:10^{23}$

$=0.5028\:\times\:10^{17}$

$\bold=5.028\:\times\:10^{16}\:molecules$

3)$\boxed{5\:\times\:{10}^{16}}$