b
moles \(=\frac{\text { number of molecules }}{6 \times 10^{23}}=\frac{\text { given mass }}{\text { molar mass }}\)

\(\Rightarrow\) molar mas \(=\frac{10 \times 6.023 \times 10^{23}}{6.023 \times 10^{22}}=100 g / mol\)

\(\Rightarrow\) molarity \(=\frac{\text { moles of solute }}{\text { volume of } \operatorname{sol}^{n}(\ell)}=\frac{(5 / 100)}{2}\)

\(=0.025\)