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 ' $x$ ' in $2 \mathrm{~L}$ solution is______________ $\times 10^{-3}$.
Solution:
(25)
Number of mole of $x=\frac{6.022 \times 10^{22}}{6.022 \times 10^{23}}=\frac{10}{\text { Molar mass of } x}$
So molar mass of $x=100 \mathrm{~g}$
Molarity $=\frac{5}{100 \times 2}=0.025 \mathrm{M}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.