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Pls solving the follin

qsnImg
Asked by ntg432000 28th April 2019, 4:25 PM
Answered by Expert
Answer:
Q. 79.
 
Given:
 
Mass = 15 g
 
Volume = 500 ml
 
We know,
 
straight M space equals space straight pi over RT

space space space equals fraction numerator 6.41 over denominator 0.082 cross times 298 end fraction

space space space equals space 0.262 space straight M
 
Therefore, 0.262 mol of particles present in 1 litre of solution and 0.131 mol of particles present in 500 ml of solution.
 
Each mole of sucrose gives 1 mol of particles.
Each mole of NaCl gives 2moles of particals.
 
Consider,
Mass of C12H22O11 be X g and
 
Mass of NaCl is (15 - X) g 
 
No space of space moles space equals space fraction numerator Mass space over denominator Molar space mass end fraction

space space space space space space space space 0.131 space space equals space fraction numerator straight X over denominator 342.3 end fraction plus space fraction numerator 2 open parentheses 15 minus straight X close parentheses over denominator 58.44 end fraction



space space space space space space space space space space space 0.131 space space equals fraction numerator space 58.44 open parentheses straight X close parentheses space plus space space open parentheses 342.3 close parentheses cross times 2 open parentheses 15 minus straight X close parentheses space space space space over denominator 342.3 cross times 58.44 end fraction space

space space space space space space space space space space space 0.131 space space equals space space fraction numerator 58.44 straight X space plus 10269 space minus space 684.6 straight X over denominator 20004.01 end fraction

space space space space space space space space space space 2620.5 space equals space minus 626.16 straight x plus 10269 space

space space space space space space space minus 7648.5 space equals space minus 626.16 straight x

space space space space space space space space space space space space space space space space space straight X space equals space fraction numerator 7648.5 over denominator 626.16 end fraction

space space space space space space space space space space space space space space space space straight X space equals space 12.21 space straight g

space space space space space therefore space Sucrose space equals space 12.21 space straight g

Pecentage space of space sucrose comma

space space space space space space space space space space space space space space fraction numerator 12.2 over denominator 15 end fraction cross times 100
space space
space space space space space space space space space space space equals space 81.42 space percent sign space sucrose

For space NaCl comma

100 space minus 81.42

equals space 18.57 space percent sign space of space NaCl
space space 
 
Pertange composition,
 
Sucrose is 81.42 %
 
NaCl is 18.57%
Answered by Expert 29th April 2019, 2:33 PM
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