Complete combustion of 3 g of ethane gives x × 1022 molecules of water. The value of x is (Round off to the Nearest Integer). [Use: N A =6.023 × 1023; Atomic masses in u: C: 12.0 ; O: 16.0 ; H: 1.0]

Question

Complete combustion of 3 g of ethane gives x × 1022 molecules of water. The value of x is (Round off to the Nearest Integer). [Use: N A =6.023 × 1023; Atomic masses in u: C: 12.0 ; O: 16.0 ; H: 1.0] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/851f9e4a8320580bb11431f8dc65bb3e-.png" /> No. of molecules =0.3 × 6.023 × 1023 =18.069 × 1022

Q. Complete combustion of $3 g$ of ethane gives $x \times 1 0^{22}$ molecules of water. The value of $x$ is (Round off to the Nearest Integer).
[Use : $N_{A} = 6.023 \times 1 0^{23}$ ; Atomic masses in $u : C : 12.0; O : 16.0; H : 1.0$ ]

No. of molecules $= 0.3 \times 6.023 \times 1 0^{23}$

$= 18.069 \times 1 0^{22}$

Q. Complete combustion of 3g of ethane gives x×1022 molecules of water. The value of x is (Round off to the Nearest Integer). [Use : NA​=6.023×1023; Atomic masses in u:C:12.0;O:16.0;H:1.0]

No. of molecules =0.3×6.023×1023 =18.069×1022

Q. Complete combustion of $3 g$ of ethane gives $x \times 1 0^{22}$ molecules of water. The value of $x$ is (Round off to the Nearest Integer).
[Use : $N_{A} = 6.023 \times 1 0^{23}$ ; Atomic masses in $u : C : 12.0; O : 16.0; H : 1.0$ ]

No. of molecules $= 0.3 \times 6.023 \times 1 0^{23}$

$= 18.069 \times 1 0^{22}$