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$n_{HNO_3} = \frac{40,3.1,24.37,8}{100.63} = 0,3$ mol
$n_{KNO_3} = n_{KOH} = n_{HNO_3} = 0,3$ mol
$m_{dd KOH} = \frac{0,3.56.100}{33,6} = 50$ gam
=> $m_{dd} = m_{ddHNO_3} + m_{ddKOH} = 40,3.1,24 + 50 = 100$ gam
ở $0^oC : m_{KNO_3} = \frac{100.11,6}{100} = 11,6$ gam
=> $m_{muối tách} = 0,3.101 - 11,6 = 18,7$ gam
Ở $0^oC : 11,6$ gam tan trong $88,4$ gam $H_2O$ => tạo $100$ gam dd
$S ------- 100$ gam $H_2O$
=> $S = \frac{100.11,6}{88,4} = 13,12$ gam
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Trả lời 27-07-13 11:28 AM
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