Free residual chlorine concentration in water was measured to be $2 \mathrm{mg} / \mathrm{l}$ (as $\mathrm{Cl}_2$ ). The pH of water is 8.5. By using the chemical equation given below, the HOCl concentration (in $\mu$ moles $/ \mathrm{l}$ ) in water is $\qquad$ (round off to one decimal place).
$$ \mathrm{HOCl} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OCl}^{-}, \mathrm{pK}=7.50 $$
Atomic weight: Cl(35.5)
A compound has a general formula $\mathrm{C}_{\mathrm{a}} \mathrm{H}_{\mathrm{b}} \mathrm{O}_{\mathrm{c}} \mathrm{N}_{\mathrm{d}}$ and molecular weight 187. A $935 \mathrm{mg} / \mathrm{l}$ solution of the compound is prepared in distilled deionized water. The Total Organic Carbon (TOC) is measured as $360 \mathrm{mg} / \mathrm{l}$ (as C). The Chemical Oxygen Demand (COD) and the Total Kjeldahl Nitrogen (TKN) are determined as $600 \mathrm{mg} / \mathrm{l}$ (as $\mathrm{O}_2$ ) and $140 \mathrm{mg} /$ I (as N), respectively (as per the chemical equation given below). Which of the following ptions is/are CORRECT?
$$ \mathrm{C}_{\mathrm{a}} \mathrm{H}_{\mathrm{b}} \mathrm{O}_{\mathrm{c}} \mathrm{~N}_{\mathrm{d}}+\frac{(4 \mathrm{a}+\mathrm{b}-2 \mathrm{c}-3 \mathrm{~d})}{4} \mathrm{O}_2 \rightarrow \mathrm{aCO}_2+\frac{\mathrm{b}-3 \mathrm{~d}}{2} \mathrm{H}_2 \mathrm{O}+\mathrm{dNH}_3 $$
Atomic weight : $\mathrm{C}(12), \mathrm{H}(1), \mathrm{O}(16), \mathrm{N}(14)$
The analyses results of a water sample are given below. The non-carbonate hardness of the water (in $\mathrm{mg} / \mathrm{L}$ ) as $\mathrm{CaCO}_3$ is __________ (in integer).
$$ \begin{aligned} & \mathrm{Ca}^{2+}=150 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_3 \\ & \mathrm{Mg}^{2+}=40 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_3 \\ & \mathrm{Fe}^{2+}=10 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_3 \\ & \mathrm{Na}^{+}=50 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_3 \\ & \mathrm{~K}^{+}=10 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_3 \\ & \mathrm{CO}_3{ }^{2-}=120 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_3 \\ & \mathrm{HCO}_3{ }^{-}=30 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_3 \end{aligned} $$
$\mathrm{Cl}^{-}=50 \mathrm{mg} / \mathrm{L}$ as $\mathrm{CaCO}_3$; Other anions were not analysed.