6.023 times 10^{22} molecules are present in | vedclass

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(B) Step $1$: Calculate the molar mass of substance $'x'$.$	ext{Moles} = \frac{	ext{Number of molecules}}{N_A} = \frac{6.023 	imes 10^{22}}{6.023 \

Vedclass · Accepted Answer

$25$

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(B) Step $1$: Calculate the molar mass of substance $'x'$.$	ext{Moles} = \frac{	ext{Number of molecules}}{N_A} = \frac{6.023 	imes 10^{22}}{6.023 	imes 10^{23}} = 0.1 \ mol$.$	ext{Molar mass} = \frac{	ext{Given mass}}{	ext{Moles}} = \frac{10 \ g}{0.1 \ mol} = 100 \ g/mol$.Step $2$: Calculate the molarity of the solution.$	ext{Moles of } x = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{5 \ g}{100 \ g/mol} = 0.05 \ mol$.$	ext{Molarity} (M) = \frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{0.05 \ mol}{2 \ L} = 0.025 \ M$.Step $3$: Express in terms of $10^{-3}$.$0.025 = 25 	imes 10^{-3}$.

$6.023 \times 10^{22}$ molecules are present in $10 \ g$ of a substance $'x'.$ The molarity of a solution containing $5 \ g$ of substance $'x'$ in $2 \ L$ solution is.......... $\times 10^{-3}$

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