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Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Problem –
Equivalent Weight
• Calculate the equiv. wt. (g/equiv.)
• of ammonia NH3 ?
Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Problem - Moles
Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Problem-Molarity
•Example:
•Determine the Molarity of 1.05g
Formaldehyde CH2O that dissolved in
250mL water ?
Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Problem- Normality
•Problem
Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Problem
-ppt, ppm, ppb for Solid Samples
Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Problem
Concentration in (mequiv/L)
Application of Statistical and mathematical equations in Chemistry Part 1
Application of Statistical and mathematical equations in Chemistry Part 1
Problem-
Percent Concentration
• Calculate the (%w/w), (%v/v), and (%w/v) for
3.0g of 10mL solute that exist in 100mL water ?
(100mL = 100g of water).
Calculate the (%w/w), (%v/v), and (%w/v) for 3.0g of 10mL
solute that exist in 100mL water ? (100mL = 100g of water).
Solution :
wt percent (%w/w) = 3g/100g = 3%
vol percent (%v/v) = 10ml/100ml = 10%
Wt.vol percent (%w/v) = 3g/100ml = 3%
wt. percent (%w/w) = ( wt. of solute (g) / wt. of solution (g) ) * 100%
vol. percent (%v/v) = ( vol. of solute (mL) / wt. of solution (mL) ) * 100%
wt. vol. percent (%w/w) = ( wt. of solute (g) / vol. of solution (mL) ) * 100%
Application of Statistical and mathematical equations in Chemistry Part 1
Problem- Density
Solution :
1.42g/cm3 = 1420g/L
no of g. per liter =1420 *70 /100=994g/L
M HNO3=499/63.0=16 mol./L

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Application of Statistical and mathematical equations in Chemistry Part 1