Question 1 |

A 100 mg of HNO_3 (strong acid) is added to water, bringing the final volume to
1.0 liter. Consider the atomic weights of H, N, and O, as 1 g/mol, 14 g/mol, and
16 g/mol, respectively. The final pH of this water is
(Ignore the dissociation of water.)

2.8 | |

6.5 | |

3.8 | |

8.5 |

Question 1 Explanation:

HNO_3\rightarrow H^++NO_3^{-}

\begin{aligned} \text{1 mole of }HNO_3&=\text{1 mole of }H^+\\ \text{No. of mole of }HNO_3&=\frac{100}{(1+14+16 \times 3)}\\ &=1.587 m\; mol\\ [H^+] \text{ in 1 lit. of water }&=1.587 m\;mol/Lit\\ &=1.587 \times 10^{-3}\;mol/Lit\\ pH&=-\log[H^+]\\ &=\log[1.587 \times 10^{-3}]\\ &=2.8 \end{aligned}

\begin{aligned} \text{1 mole of }HNO_3&=\text{1 mole of }H^+\\ \text{No. of mole of }HNO_3&=\frac{100}{(1+14+16 \times 3)}\\ &=1.587 m\; mol\\ [H^+] \text{ in 1 lit. of water }&=1.587 m\;mol/Lit\\ &=1.587 \times 10^{-3}\;mol/Lit\\ pH&=-\log[H^+]\\ &=\log[1.587 \times 10^{-3}]\\ &=2.8 \end{aligned}

Question 2 |

In a water sample, the concentrations of Ca^{2+},Mg^{2+}, and HCO_3^{-} are 100 mg/L, 36 mg/L and 122 mg/L, respectively. The atomic masses of various elements are: Ca = 40, Mg = 24, H = 1, C = 12, O = 16.

The total hardness and the temporary hardness in the water sample (in mg/L as CaCO_3) will be

The total hardness and the temporary hardness in the water sample (in mg/L as CaCO_3) will be

400 and 100, respectively. | |

400 and 300, respectively. | |

500 and 100, respectively. | |

800 and 200, respectively. |

Question 2 Explanation:

Total Hardness,

\begin{aligned} TH&=[Ca^{+2}] \times \frac{50}{20}+[Mg^{+2}] \times \frac{50}{12}\\ &=100 \times \frac{50}{20}+36 \times \frac{50}{12}\\ &=400 mg/l \text{ as } CaCO_3 \end{aligned}

Total alkalinity,

TA=[HCO_3^-]\times \frac{50}{61}=122 \times \frac{50}{61}=100\; mg/l \text{ as } CaCO_3

Temporary hardness (carbonate hardness)

=min\left\{\begin{matrix} TA= 100\; mg/l \text{ as } CaCO_3\\ TH=400\; mg/l \text{ as } CaCO_3 \end{matrix}\right.=100 \; mg/l \text{ as } CaCO_3

\begin{aligned} TH&=[Ca^{+2}] \times \frac{50}{20}+[Mg^{+2}] \times \frac{50}{12}\\ &=100 \times \frac{50}{20}+36 \times \frac{50}{12}\\ &=400 mg/l \text{ as } CaCO_3 \end{aligned}

Total alkalinity,

TA=[HCO_3^-]\times \frac{50}{61}=122 \times \frac{50}{61}=100\; mg/l \text{ as } CaCO_3

Temporary hardness (carbonate hardness)

=min\left\{\begin{matrix} TA= 100\; mg/l \text{ as } CaCO_3\\ TH=400\; mg/l \text{ as } CaCO_3 \end{matrix}\right.=100 \; mg/l \text{ as } CaCO_3

Question 3 |

A water filtration unit is made of uniform-size sand particles of 0.4 mm diameter with a shape factor of 0.84 and specific gravity of 2.55. The depth of the filter bed is 0.70 m and the porosity is 0.35. The filter bed is to be expanded to a porosity of 0.65 by hydraulic backwash. In the terminal settling velocity of sand particles during backwash is 4.5 cm/s, the required backwash velocity is

5.79 \times 10^{-3} \mathrm{~m} / \mathrm{s} | |

6.35 \times 10^{-3} \mathrm{~m} / \mathrm{s} | |

0.69 cm/s | |

0.75 cm/s |

Question 3 Explanation:

\begin{aligned} n^{\prime} &=\text { Porosity of expanded bed } \\ n^{\prime} &=\left(\frac{V_{B}}{V_{s}}\right)^{0.22} \\ 0.65 &=\left(\frac{V_{B}}{4.5 \mathrm{~cm} / \mathrm{s}}\right)^{0.22} \\ V_{B} &=6.35 \times 10^{-3} \mathrm{~m} / \mathrm{s} \end{aligned}

Question 4 |

The internal \left(d_{i}\right) and external \left(d_{o}\right) diameters of a Shelby sampler are 48 mm and 52 mm, respectively. The area ratio \left(A_{r}\right)
of the sampler (in %,round off to two decimal palces) is _____________

12.45 | |

28.56 | |

47.12 | |

17.36 |

Question 4 Explanation:

Outside diameter = 52 mm

Inside diameter= 48 mm

\begin{aligned} A_{r}&=\frac{\frac{\pi}{4}\left(D_{2}\right)^{2}-\frac{\pi}{4}\left(D_{1}\right)^{2}}{\frac{\pi}{4}\left(D_{1}\right)^{2}} \times 100 \\ A_{r}&=\frac{(52)^{2}-(48)^{2}}{(48)^{2}} \times 100=17.36 \% \end{aligned}

Inside diameter= 48 mm

\begin{aligned} A_{r}&=\frac{\frac{\pi}{4}\left(D_{2}\right)^{2}-\frac{\pi}{4}\left(D_{1}\right)^{2}}{\frac{\pi}{4}\left(D_{1}\right)^{2}} \times 100 \\ A_{r}&=\frac{(52)^{2}-(48)^{2}}{(48)^{2}} \times 100=17.36 \% \end{aligned}

Question 5 |

A water sample is analyzed for coliform organisms by the multiple-tube fermentation method. The results of confirmed test are as follows:

\begin{array}{|l|c|c|} \hline \text { Sample size (mL) } & \begin{array}{c} \text { Number of positive } \\ \text { results out of } 5 \text { tubes } \end{array} & \begin{array}{c} \text { Number of negative } \\ \text { results out } 5 \text { tubes } \end{array} \\ \hline 0.01 & 5 & 0 \\ \hline 0.001 & 3 & 2 \\ \hline 0.0001 & 1 & 4 \\ \hline \end{array}

The most probable number (MPN) of coliform organisms for the above results is to be obtained using the following MPN Index.

The MPN of coliform organisms per 100 mL is

\begin{array}{|l|c|c|} \hline \text { Sample size (mL) } & \begin{array}{c} \text { Number of positive } \\ \text { results out of } 5 \text { tubes } \end{array} & \begin{array}{c} \text { Number of negative } \\ \text { results out } 5 \text { tubes } \end{array} \\ \hline 0.01 & 5 & 0 \\ \hline 0.001 & 3 & 2 \\ \hline 0.0001 & 1 & 4 \\ \hline \end{array}

The most probable number (MPN) of coliform organisms for the above results is to be obtained using the following MPN Index.

The MPN of coliform organisms per 100 mL is

1100000 | |

110000 | |

1100 | |

110 |

Question 5 Explanation:

The sample size is 0.01 ml, 0.001 ml and 0.0001 ml which is 1000 times lesser than
the standard of 10 ml, 1 ml and 0.1 ml.

The positive set is 5-3-1 and w.r.t. the positive combination,

MPN/100 ml will be 110 x 1000 = 110000

The positive set is 5-3-1 and w.r.t. the positive combination,

MPN/100 ml will be 110 x 1000 = 110000

Question 6 |

Which one of the following is correct?

The partially treated effluent from a food processing industry, containing high concentration of biodegradable organics, is being discharged into a flowing river at a point P. If the rate of degradation of the organics is higher than the rate of aeration, then dissolved oxygen of the river water will be lowest at point P. | |

The most important type of species involved in the degradation of organic matter in the case of activated sludge process based wastewater treatment is chemoheterotrophs | |

For an effluent sample of a sewage treatment plant, the ratio BOD 5-\text { day }, 20^{\circ} C upon ultimate BOD is more than 1 | |

A young lake characterized by low nutrient content and low plant productivity is called eutrophic lake |

Question 6 Explanation:

Chemoheterotrophs are the organisms which generate energy with the help of chemical reactions by consuming organic matter during their metabolism. These are the species which help to degrade the organic matter in ASP.

Question 7 |

A sample of water contain an organic compound C_8H_{16}O_8 at a concentration of 10^{-3} mol/litre.
Given that the atomic weight of C = 12 g/mol, H = 1 g/mol, and O = 16 g/mol, the
theoretical oxygen demand of water (in g of O_2 per litre, round off to two decimal places),
is ___________

0.52 | |

0.25 | |

0.82 | |

0.64 |

Question 7 Explanation:

C_8H_{16}O_8 of conc 10^{-3} moles/lt required O_2 (in gm/lt)

C_8H_{16}O_8 +8O_2\rightarrow 8CO_2+8H_2O

1 mole of C_8H_{16}O_8 requires 8 moles of O_2 for its decomposition

\begin{aligned} 240 gm &= 128 gm\\ 10^{-3} moles&=8 \times 10^{-3}moles \\ &= 8 \times 10^{-3} \times 32\\ &= 0.256 gm/lt \end{aligned}

C_8H_{16}O_8 +8O_2\rightarrow 8CO_2+8H_2O

1 mole of C_8H_{16}O_8 requires 8 moles of O_2 for its decomposition

\begin{aligned} 240 gm &= 128 gm\\ 10^{-3} moles&=8 \times 10^{-3}moles \\ &= 8 \times 10^{-3} \times 32\\ &= 0.256 gm/lt \end{aligned}

Question 8 |

The ion product of water (pK_w) is 14. If a rain water sample has a pH of 5.6, the
concentration of OH^- in the sample (in 10^{-9} mol/litre, round off to one decimal place),
is ________

3.98 | |

8.42 | |

6.24 | |

7.46 |

Question 8 Explanation:

\begin{aligned} pH+pOH&=14\\ pOH&=14-5.6=8.4\\ -log[OH^-]&=8.4\\ [OH^-]&=10^{-8.4}\; moles/lt\\ &=10^{-8.4+9} \times 10^{-9}\; moles/lt\\ &=3.98 \times 10^{-9} \; moles/lt \end{aligned}

Question 9 |

Alkalinity of water, in equivalent/litre (eq/litre), is given by

\{HCO_3^-\}+2\{CO_3^{2-}\}+\{OH^-\}-\{H^+\}

where, { } represents concentration in mol/litre. For a water sample, the concentration of HCO_3^? = 2 \times 10^{-3} mol/litre, CO_3^{2-} = 3.04 \times 10^{-4} mol/litre and the pH of water = 9.0. The atomic weights are : Ca = 40; C = 12; and O = 16. If the concentration of OH^- and H^+ are NEGLECTED, the alkalinity of the water sample (in mg/litre as CaCO_3), is

\{HCO_3^-\}+2\{CO_3^{2-}\}+\{OH^-\}-\{H^+\}

where, { } represents concentration in mol/litre. For a water sample, the concentration of HCO_3^? = 2 \times 10^{-3} mol/litre, CO_3^{2-} = 3.04 \times 10^{-4} mol/litre and the pH of water = 9.0. The atomic weights are : Ca = 40; C = 12; and O = 16. If the concentration of OH^- and H^+ are NEGLECTED, the alkalinity of the water sample (in mg/litre as CaCO_3), is

130.4 | |

100 | |

50 | |

65.2 |

Question 9 Explanation:

Alkalinity of water sample is due to presence of

[HCO_3^-] and [HCO_3^{2-}]

Total alkalinity = 1 mole of [HCO_3^-] + 2 mole of [HCO_3^{2-}] in terms of CaCO_3

=(2 \times 10^{-3} \times 50 +2 \times 3.04 \times 10^{-4} \times 50) \times 10^3 mg/l

=130.4 mg/l \; as \; CaCO_3

[HCO_3^-] and [HCO_3^{2-}]

Total alkalinity = 1 mole of [HCO_3^-] + 2 mole of [HCO_3^{2-}] in terms of CaCO_3

=(2 \times 10^{-3} \times 50 +2 \times 3.04 \times 10^{-4} \times 50) \times 10^3 mg/l

=130.4 mg/l \; as \; CaCO_3

Question 10 |

A river has a flow of 1000 million litres per day (MLD), BOD_5 of 5 mg/litre and Dissolved
Oxygen (DO) level of 8 mg/litre before receiving the wastewater discharge at a location.
For the existing environmental conditions, the saturation DO level is 10 mg/litre in the
river. Wastewater discharge of 100 MLD with the BOD_5 of 200 mg/litre and DO level of
2 mg/litre falls at that location. Assuming complete mixing of wastewater and river water,
the immediate DO deficit (in mg/litre, round off to two decimal places), is _________.

7.45 | |

2.54 | |

1.55 | |

4.45 |

Question 10 Explanation:

\begin{aligned} DO_{mix}&=\frac{DO_{s}Q_s+DO_RQ_R}{Q_s+Q_R}\\ &=\frac{2 \times 100+8 \times 1000}{100+1000}\\ &=7.45 \; \text{mg/l}\\ DO&=DO_{sat}-DO_{mix}\\ &=10-7.45=2.545 \; \text{mg/l} \end{aligned}

There are 10 questions to complete.

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