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

 $\mathrm{Molarmass} \mathrm{of} \text{'}\mathrm{x}\text{'}=\frac{10}{6.023\times {10}^{22}}\times \left(6.023\times {10}^{23}\right)=100\mathrm{g}/\mathrm{mol}$

 $\mathrm{MM}=100 \mathrm{g}/\mathrm{mol}$

Molarity $=\frac{\frac{5}{100}}{2}=0.025=25\times {10}^{-3}M$

As a result, the solution's molarity, which is 5g of substance "x" in 2 L of solution, is $25$.