Answer to Question #300597 in General Chemistry for Justin

Question #300597

 If 7.5 g of sodium nitrate, NaNO3 is dissolved in 85ml of H2O (density of H2O = 1.0g/ml), calculate the %(wt/wt) concentration of H2O in the solution. 


Given:


 Solution:



 A brand of rubbing alcohol says, it contains 70% (vol/vol) isopropyl alcohol. How many mL of isopropyl alcohol are there in 600 ml of the solution in the bottle? 

Given:


 Solution:



 The label of betadine skin cleanser says 7.5 % solution. Taking it to be % (wt/vol), how many grams of betadine (Providone-Iodine) are present in 50 mL bottle? 

Given:


 Solution:



 Calculate the mole fraction of sulfuric acid (H2SO4) in 8% (wt/wt) aqueous H2SO4 solution (molar masses: H2SO4 = 98 g/mol, H2O = 18 g/mol 

Given:


 Solution:



 Calculate the molar concentration of a solution that contains 23 g of potassium hydroxide. KOH in 250 ml of solution. Molar mass of KOH is 56 g/mol. 

Given: 


Solution:




1
Expert's answer
2022-02-22T14:31:03-0500

Part 1


Mass of water "=85\u00d71=85g"

"\\%" concentration of water "=\\frac{85}{(85+7.5)}\u00d7100\\%=91.89\\%"


Part 2


Ml of isopropyl alcohol "=\\frac{70}{100}\u00d7600=420ml"


Part 3

"\\frac{Mass}{50}=\\frac{7.5}{100}\\implies" mass "=3.75g"


Part 4

Mass of H2SO4 "=8g"

Mass of water "=100-8=92g"


Moles of H2SO4 "=\\frac{8}{98}=0.08163"

Moles of water "=\\frac{92}{18}=5.11111"


Mole fraction "=\\frac{0.08163}{(5.11111+0.08163)}=0.01572"



Part 5

Mole concentration "=\\frac{23}{56}\u00d7\\frac{1000}{250}"

"=1.643M"









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