Question

molecule wala question...please mujhe ye bhi bataiyega ki ise kese karte hai ​

Answer 1

Answer:

$c) \: 1g \: H_2 \:$

Explanation:

At first we need to Calculate the number of molecules in each of the following options -

$1g \: CO_2 \\ \\ CO_2 \: = 12 + 16 \times 2 = 12 + 32 \\ = 44 \\ \\ mass \: = mole(n) \times gram \: molecular \: mass \\ \\ mole(n) = \frac{mass}{gram \: molecular \: mass} \\ \\ = \frac{1}{44} mole \\ \\ now \\ \\ number \: of \: molecules \: = mole(n) \times N_{A} \\ \\ \frac{1}{44} \times 6 \times {10}^{23} \\ \\ \\ 1g \: O_2 \\ \\ O_2 = 16 \times 2 = 32 \\ \\ mole = \frac{1}{32} \\ \\ number \: of \: molecule \: = \frac{1}{32} \times 6 \times {10}^{23} \\ \\ 1g \: \: H_2 \\ \\ \: H_2 \: = 1 \times 2 = 2 \\ \\ mole = \frac{1}{2} \\ \\ number \: of \: molecules = \frac{1}{2} \times 6 \times {10}^{23} \\ \\ = 3 \times {10}^{23} \\ \\ \\ 1g \: CH_4 \: \\ \\ CH_4 \: = 12 + 1 \times 4 = 12 + 4 \\ = 16 \\ \\ mole = \frac{1}{16} \\ \\ number \: of \: molecule = \frac{1}{16} \times 6 \times {10}^{23} \\ \\ \: here \: we \: take \: approx \: value \: of \: N_{A} \\ correct \: value \: is \: 6.022 \times {10}^{23} \\ \\ hence \\ \\ 1g \: \: of \: H_2 \: is \: correct \: answer$

Answer 2

Answer:

Correct option is (c) 1gH2