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: $6.023 	imes 10^{22}$ molecules correspond to $10 \ g$.Using the formula: $	ext{Moles

Vedclass · Accepted Answer

$25$

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

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

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