Question

The total number of valence electron in 1.9 g of H2O+ion is (NA = Avogadro's number)
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Answer 1

Answer:

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Answer 2

Answer:

The total number of valence electrons in $1.9g$ of $H_{2}O^{+}$ ions calculated is $4.21\times 10^{23} electrons$.

Explanation:

Data given,

The given mass of the $H_{2}O^{+}$ ions = $1.9g$

The total number of the valence electrons in $1.9g$ of $H_{2}O^{+}$ ions =?

Firstly, we have to calculate the number of moles in $1.9g$ of $H_{2}O^{+}$ ions.

As we know,

• The molar mass of water ( $H_{2}O$ ) = $18g/mol$.

Therefore,

• The number of moles of the $H_{2}O^{+}$ ions = $\frac{Given mass}{Molar mass}$ = $\frac{1.9}{18}$ = $0.10mol$

Also,

• $1 mol = 6.022\times 10^{23}ions$

So,

• $0.10mol = ( 0.10\times 6.022\times 10^{23}) ions$ = $6.022\times 10^{22} ions$

Now, we have to calculate the number of valence electrons in one $H_{2}O^{+}$ ion.

• The valence electrons in $H_{2}O^{+}$ = ( 2 × valence electron in hydrogen ) + ( 1 × valence electrons in oxygen ) - cationic charge
• The valence electrons in $H_{2}O^{+}$ = $(2\times 1) + (1\times 6) -1$ = $7electrons$.

Hence, the number of valence electrons in $6.022\times 10^{22} ions$ = $(7\times 6.022\times 10^{22} )electrons$ = $4.21\times 10^{23} electrons$.

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