1.0 g of has same number of molecules as in [2024]

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Solution

Q.
1.0 g of $H_{2}$ has same number of molecules as in [2024]

1 14 g of $N_{2}$

2 18 g of $H_{2} O$

3 16 g of CO

4 28 g of $N_{2}$

Ans.
(1)

Amount Number of molecules

1.0 g of $H_{2}$ $\frac{N_{A}}{2} = 3.01 \times 10^{23}$

14 g of $N_{2}$ $\frac{N_{A}}{2} = 3.01 \times 10^{23}$

18 g of $H_{2} O$ $N_{A} = 6.022 \times 10^{23}$

16 g of CO $\frac{4 N_{A}}{7} = 3.44 \times 10^{23}$

28 g of $N_{2}$ $N_{A} = 6.022 \times 10^{23}$

Amount	Number of molecules
1.0 g of $H_{2}$	$\frac{N_{A}}{2} = 3.01 \times 10^{23}$
14 g of $N_{2}$	$\frac{N_{A}}{2} = 3.01 \times 10^{23}$
18 g of $H_{2} O$	$N_{A} = 6.022 \times 10^{23}$
16 g of CO	$\frac{4 N_{A}}{7} = 3.44 \times 10^{23}$
28 g of $N_{2}$	$N_{A} = 6.022 \times 10^{23}$

Solution

Q. 1.0 g of H2 has same number of molecules as in [2024]

1 14 g of N2

2 18 g of H2O

3 16 g of CO

4 28 g of N2

Ans. (1) Amount Number of molecules 1.0 g of H2 NA2=3.01×1023 14 g of N2 NA2=3.01×1023 18 g of H2O NA=6.022×1023 16 g of CO 4NA7=3.44×1023 28 g of N2 NA=6.022×1023

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Q.
1.0 g of $H_{2}$ has same number of molecules as in [2024]

1 14 g of $N_{2}$

2 18 g of $H_{2} O$

4 28 g of $N_{2}$