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

There are 5 questions to complete.

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