1) 2) Calculate the MW of the following: i) (NH4)3PO4, ii) K2HPO4, iii) Ca(OCI)2 (N-14, H-1, P-31, K-39, O-16, Ca-40, Cl-35-5) Find out the mole: i) 2500 mg of NH4NO3 (N-14, O-16), ii) 0.751 L of ethyl formate (Sp. gv. -1.00124), (C-12, O-16, H-1), iii) 0.0889 Kg of Al2SiO5 (Al-27, Si-14, O-16) 3) What is the conc (M) when? 4) i) 295 gm of KMnO4 dissolves in 2.002 L of water. (K-39, Mn- 25) ii) 0.0065 Kg of KH2PO4 dissolves in 0.2305 mL water. (P-31, 0-16) How much (gm) solute is dissolved in? i) 2100 mL 0.025 M solution of (NH4)2SO4 ii) 0.399 L 1.005 M solution of C16H31KO2 5) 6) How much total glucose is present in a 75 Kg female person? (a normal person (70 kg BW) has approximately 43L body fluid). Please express the normal salt (NaCl) concentration in body fluid into molarity (mM). 7) Commercial fuming Sulphuric acid (Oleum-H2S206) is 99.9%. solution. Please convert it into molarity. 8) Find out the Volume (dm3) of product (gas) at RTP when 0.58 M, 150 mL NaOH (aq.) reacts with 350 mL, 0.25 NH4Cl. 9) The above reaction has the product Ammonia, which when dissolved in 650 mL ethanol makes an alkaline ethanolic solution. Find its molarity (M) 10) Calculate the adult dose as per the BW of the baby. (Child dose- 50 mg and the BW of the baby is 48 lb (British pound) (1

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Please solve question number 9.

1)
2)
Calculate the MW of the following:
i) (NH4)3PO4, ii) K2HPO4, iii) Ca(OCI)2
(N-14, H-1, P-31, K-39, O-16, Ca-40, Cl-35-5)
Find out the mole:
i) 2500 mg of NH4NO3 (N-14, O-16), ii) 0.751 L of ethyl
formate (Sp. gv. -1.00124), (C-12, O-16, H-1), iii) 0.0889 Kg
of Al2SiO5 (Al-27, Si-14, O-16)
3) What is the conc (M) when?
i) 295 gm of KMnO4 dissolves in 2.002 L of water. (K-39, Mn-
25)
ii) 0.0065 Kg of KH2PO4 dissolves in 0.2305 mL water. (P-31,
0-16)
4) How much (gm) solute is dissolved in?
i) 2100 mL 0.025 M solution of (NH4)2SO4
ii) 0.399 L 1.005 M solution of C16H31KO2
How much total glucose is present in a 75 Kg female person?
(a normal person (70 kg BW) has approximately 43L body
fluid).
6) Please express the normal salt (NaCl) concentration in body
fluid into molarity (mM).
7)
Commercial fuming Sulphuric acid (Oleum-H2S206) is 99.9%.
solution. Please convert it into molarity.
Find out the Volume (dm3) of product (gas) at RTP when 0.58
M, 150 mL NaOH (aq.) reacts with 350 mL, 0.25 NH4Cl.
The above reaction has the product Ammonia, which when
dissolved in 650 mL ethanol makes an alkaline ethanolic
solution. Find its molarity (M)
10) Calculate the adult dose as per the BW of the baby. (Child dose-
50 mg and the BW of the baby is 48 lb (British pound) (1
lb=0.453 Kg)
Transcribed Image Text:1) 2) Calculate the MW of the following: i) (NH4)3PO4, ii) K2HPO4, iii) Ca(OCI)2 (N-14, H-1, P-31, K-39, O-16, Ca-40, Cl-35-5) Find out the mole: i) 2500 mg of NH4NO3 (N-14, O-16), ii) 0.751 L of ethyl formate (Sp. gv. -1.00124), (C-12, O-16, H-1), iii) 0.0889 Kg of Al2SiO5 (Al-27, Si-14, O-16) 3) What is the conc (M) when? i) 295 gm of KMnO4 dissolves in 2.002 L of water. (K-39, Mn- 25) ii) 0.0065 Kg of KH2PO4 dissolves in 0.2305 mL water. (P-31, 0-16) 4) How much (gm) solute is dissolved in? i) 2100 mL 0.025 M solution of (NH4)2SO4 ii) 0.399 L 1.005 M solution of C16H31KO2 How much total glucose is present in a 75 Kg female person? (a normal person (70 kg BW) has approximately 43L body fluid). 6) Please express the normal salt (NaCl) concentration in body fluid into molarity (mM). 7) Commercial fuming Sulphuric acid (Oleum-H2S206) is 99.9%. solution. Please convert it into molarity. Find out the Volume (dm3) of product (gas) at RTP when 0.58 M, 150 mL NaOH (aq.) reacts with 350 mL, 0.25 NH4Cl. The above reaction has the product Ammonia, which when dissolved in 650 mL ethanol makes an alkaline ethanolic solution. Find its molarity (M) 10) Calculate the adult dose as per the BW of the baby. (Child dose- 50 mg and the BW of the baby is 48 lb (British pound) (1 lb=0.453 Kg)
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