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calculate the molality of 1L soln of 93% h2so4 (w/v) the density of soln is 1.84 g/mL

Asked by soha 11th August 2012, 10:03 PM
Answered by Expert
Answer:
93 % H2SO4 w/v means 93 g of H2SO4 is present in 100cc of solution.
 
Since volume is 1000ml, amount of H2SO4 is 930 g/litre. Also density is given 1.84g/ml
M= density x volume = 1.84 x 1000 = 1840 g
 
wt of solvent = 1840 - 930 = 910 g
 
So, molality = 930/98 x 1000/910 = 10.43
Answered by Expert 13th August 2012, 9:23 AM
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