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

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(B) Step $1$: Calculate the molar mass of substance '$X$'.Given that $6.023 	imes 10^{22}$ molecules are present in $10 	ext{ g}$ of '$X$'.Using the

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$25$

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

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

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