tính nồng độ phần trăm của những dung dịch sau:a,20g KClrong 600g dung dịch.b,75g K2SO4 trong 1500 dung dịch.c,hòa tan 15g NaCl vào 45g nước.d,hòa tan
Question
Kennedy
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a,c%= $\frac{mct}{mdd}$*100
=$\frac{20}{600}$*100=3.3%
b,c%=$\frac{mct}{mdd}$*100
=$\frac{75}{1500}$*100=5%
c,m dd=mct+mdm
=15+45=60(g)
=>C%= $\frac{15}{60}$*100=25 %
d,vH2=4.48(litf)=>nH2=$\frac{4.48}{22.4}$=0.2(mol)
=>mh2=0.2*2=0.4(g)
=>mdd=0.4+500=500.4(g)
=>C%=$\frac{0.4}{500.4}$*100=0.08%
a) \(C{\% _{KCl}} = \frac{{20}}{{600}} = 3,33\% \)
b) \(C{\% _{{K_2}S{O_4}}} = \frac{{75}}{{1500}} = 5\% \)
c) \({m_{dd}} = {m_{NaCl}} + {m_{{H_2}O}} = 15 + 45 = 60{\text{ gam}} \to {\text{C}}{{\text{\% }}_{NaCl}} = \frac{{15}}{{60}} = 25\% \)
d) \({n_{HCl}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}} \to {{\text{m}}_{dd}} = 7,3 + 500 = 507,3{\text{ gam}} \to {\text{C}}{{\text{\% }}_{HCl}} = \frac{{7,3}}{{507,3}} = 1,44\% \)