Question:
$6.023 \times 10^{22}$ molecules are present in $10 \mathrm{~g}$ of a substance ' $x$ '. The molarity of a solution containing $5 \mathrm{~g}$ of substance ' $\mathrm{x}$ ' in $2 \mathrm{~L}$ solution is $\times 10^{-3}$.
Solution:
moles $=\frac{\text { number of molecules }}{6 \times 10^{23}}=\frac{\text { given mass }}{\text { molar mass }}$
$\Rightarrow$ molar mas $=\frac{10 \times 6.023 \times 10^{23}}{6.023 \times 10^{22}}=100 \mathrm{~g} / \mathrm{mol}$
$\Rightarrow$ molarity $=\frac{\text { moles of solute }}{\text { volume of sol }^{\mathrm{n}}(\ell)}=\frac{(5 / 100)}{2}$
$=0.025$
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