Equal moles of ${\mathrm{H}}_2\mathrm{O}$ and $\mathrm{NaCl}$ are present in a solution. Hence, molality of $\mathrm{NaCl}$ solution is

Decreasing order of mass of pure $\mathrm{NaOH}$ in each of the aqueous solution is

(I)  $50 \mathrm{g}$ of $40\%$ $(\mathrm{W}/\mathrm{W} ) \mathrm{NaOH}$
(II) $50 \mathrm{ml}$ of $50\% (\mathrm{W}/\mathrm{V}) \mathrm{NaOH} \left({\mathrm{d}}_{\mathrm{sol}}=1.2 \mathrm{g}/\mathrm{ml}\right)$
(III) $50 \mathrm{g}$ of $15 \mathrm{M} \mathrm{NaOH} \left({\mathrm{d}}_{\mathrm{sol}}=1 \mathrm{g}/\mathrm{ml}\right)$

How many moles of Lead $\left(\mathrm{II}\right)$ Chloride will be formed from a reaction between $6.5\mathrm{g}$ of $\mathrm{PbO}$ and $3.2\mathrm{g}$ of $\mathrm{HCl}$

Given the atomic mass of $\left(\mathrm{Pb}=207.2\right)$