1- What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol in 85.0 g of water ? Ans. 6%

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QUESTIONS
1- What is the weight percent of ethanol in a solution made by dissolving 5.3 g of
´cthanol in 85.0 g of water ?
Ans. 6%
2- How would you make 250 g of a 7.5% solution of glucose in water ? Ans. 18.75g
of glucose in 231.25g H;O
3- A sample of a solution weighing 850.0 g is known to contain 0.223 moles of
potassium chloriċe. What is the weight percent of potassium chloride in the
solution? Ans. 1.969%
4- Conc. HCl has a density of 1.19 g.mL and is about 36.5% by weight HCl. The
Molarity and normality of the solution is:
(a) 11.9 and 11.9
(c) 12.07 and 12.07 (d) 5.95 and 11.90 .... Show your calculations Ans. (a)
(b) 11.9 and 23.8
Answer :
Assume we have 1L of HCI
Density = Weight/ volume
Weight= 1000 x 1.19= 1190 g
Weight of HCl in solution = (1190 x 36.5) /100 = 434.35g
Molecular weight of HCI = 35.5+ 1= 36.5 g/ mol.
Number of HCI moles = 434.35 / 36.5 = 11.9 moles
Molarity of HCI =it's Normality
The answer is (a) 11.9 and 11.9
5- What is the molar concentration of a solution made by mixing 300 mL of 0.0200
M H;SO, with 200 mL of 0.0300 M H2SO,?
Ans. 0.024M
6- The mole fraction of CHs in a solution containing 37.5 g of camphor in 39.0 g
of benzene was 0.667. The molecu!ar weight of camphor is found to be:
(а) 37.5
(b) 150
(c) 112.5 (d) 300 .... Show your calculations
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Transcribed Image Text:* ZAIN IQ In. اسئلة وحلول.pdf Scanned by CamScanner QUESTIONS 1- What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ´cthanol in 85.0 g of water ? Ans. 6% 2- How would you make 250 g of a 7.5% solution of glucose in water ? Ans. 18.75g of glucose in 231.25g H;O 3- A sample of a solution weighing 850.0 g is known to contain 0.223 moles of potassium chloriċe. What is the weight percent of potassium chloride in the solution? Ans. 1.969% 4- Conc. HCl has a density of 1.19 g.mL and is about 36.5% by weight HCl. The Molarity and normality of the solution is: (a) 11.9 and 11.9 (c) 12.07 and 12.07 (d) 5.95 and 11.90 .... Show your calculations Ans. (a) (b) 11.9 and 23.8 Answer : Assume we have 1L of HCI Density = Weight/ volume Weight= 1000 x 1.19= 1190 g Weight of HCl in solution = (1190 x 36.5) /100 = 434.35g Molecular weight of HCI = 35.5+ 1= 36.5 g/ mol. Number of HCI moles = 434.35 / 36.5 = 11.9 moles Molarity of HCI =it's Normality The answer is (a) 11.9 and 11.9 5- What is the molar concentration of a solution made by mixing 300 mL of 0.0200 M H;SO, with 200 mL of 0.0300 M H2SO,? Ans. 0.024M 6- The mole fraction of CHs in a solution containing 37.5 g of camphor in 39.0 g of benzene was 0.667. The molecu!ar weight of camphor is found to be: (а) 37.5 (b) 150 (c) 112.5 (d) 300 .... Show your calculations Scanned by CamScanner
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